LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 


Class 


REINFORCED   CONCRETE 


Published   by  the 

McGraw-Hill   Boole  Company 

Vor-lc 


Succe-s-sons  to  the  Book  Departments  of  the") 

McGraw  Publishing  Company  Hill  Publishing  Company 

Publishers   of  Books  for 

Electrical  World  The  Engineering  and  Mining  Journal 

Engineering  Record  R>wer  and  The  Engineer 

Electric  Railway  Journal  American   Machinist 

Metallurgical  and  Chemical  Engineering; 


REINFORCED 
CONCRETE 

MECHANICS  AND  ELEMENTARY 

DESIGN 


BY 

JOHN    P.  BROOKS 

Director  Clarkson  School  of  Technology 

Formerly  Associate  Professor  of  Civil  Engineering 

University  of  Illinois 


McGRAW-HILL  BOOK  COMPANY 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.G. 

1911 


Copyright,  1911,  BY  THE  McGRAW-HiLL  BOOK  COMPANY 


THE  •  PLIMPTON  •  PRESS  •  NORWOOD  •  MASS  •  TJ  •  S  •  A 


PREFACE 

THIS  volume  is  designed  primarily  to  supplement  the 
usual  college  work  in  mechanics  and  masonry  design. 
With  this  end  in  view  there  is  herein  no  duplication  of 
these  subjects.  The  reader  is  referred  to  sources  of  infor- 
mation regarding  the  results  of  tests  on  reinforced  concrete 
material  and  only  such  quotations  are  given  as  serve  to 
illustrate  principles.  The  details  of  reinforced  concrete 
construction  are  constantly  changing  and  the  latest  designs 
are  to  be  found  in  the  engineering  periodicals;  conse- 
quently, matter  of  this  character  is  not  given. 

As  a  guide  to  the  selection  of  proper  constants  in  design- 
ing, much  of  the  report  of  the  "Joint  Committee"  is  given 
without  change,  and  frequent  references  to  the  same  are 
made  throughout  the  book. 

The  nomenclature  is,  usually,  made  up  of  initials  of  the 
words  indicated,  and  for  this  reason  it  was  thought  best  to 
use  St  rather  than  /«  for  the  tensile  stress  in  the  steel. 
In  general,  the  nomenclature  is  that  in  common  use. 

Several  designs  of  reinforced  concrete  structures  are 
worked  out  in  detail  with  particular  reference  to  the  proper 
sequence  of  computation.  The  principles  of  economy  in 
design  are  set  forth  and  the  diagrams  in  use  lead  to  the 
proper  selection  of  steel  and  concrete  dimensions. 

It  is  hoped  that  the  book  may  enable  the  reader  or 
student  to  become  familiar  with  the  methods  of  analysis 
and  design  of  reinforced  concrete  structures  with  as  little 
unnecessary  work  as  possible. 

J.  P.  B. 

POTSDAM,  NEW  YORK 

September,  1911. 


224422 


CONTENTS 


CHAPTER  I 

PAGE 

HISTORICAL  SKETCH 1 

Very  Early  Types  of  Reinforcement.  Development  in  Europe 
and  in  America.  Production  of  Portland  and  Natural  Cement  in 
America.  Early  Failures  of  Reinforced  Concrete  Buildings. 
Variety  of  Structures  built  of  Reinforced  Concrete.  Economy 
of  Reinforced  Concrete  Structures. 

CHAPTER  II 

THE  COMPONENT  PARTS  .  6 


CEMENT. 

Definitions.  Specific  Gravity.  Fineness.  Time  of  Setting. 
Tensile  Strength.  Constancy  of  Volume.  Sulphuric  Acid  and 
Magnesia. 

SAND. 

Specifications.  Influence  upon  Strength  of  Mortar.  United 
States  Geological  Survey  Report.  Sieve  Analysis.  Influence  of 
Weight  of  Sand.  Simple  Tests  for  Sand. 

THE  AGGREGATE. 

Specifications.     The  Water.     The  Steel. 

CHAPTER  III 

TESTS  OF  REINFORCED  CONCRETE 13 

Sources  of  Information.  Tensile  Tests.  Tension  in  Plain  Con- 
crete Beams.  Compressive  Tests.  Table  of  Compressive  Strength 
of  Concrete.  Shearing.  Definitions.  Tables  of  Shearing  Tests. 
Tests  of  Elasticity  of  Concrete.  Tables.  The  Stress-Strain  Dia- 
gram. Discussion.  Tests  of  Bond.  Length  of  Grip  to  Develop 
Tensile  Strength.  Flexural  Tests.  The  Neutral  Axis.  Deter- 
mination of  Neutral  Axis  by  Measurement.  Shifting  of  Neutral 
Axis  under  Stresses.  Beam  Failures,  by  Compression,  by  Tension, 
and  by  Diagonal  Tension.  Bond  and  Shear.  Repetition  of  Stresses. 
Expansion  and  Contraction.  Weight  of  Reinforced  Concrete. 

vii 


viii  CONTENTS 

PAGE 

The  Reinforcing  Metal.  Qualities  of  Steel.  Specifications  for 
Steel.  Proportioning  Concrete.  Quantities  in  a  Cubic  Yard  of 
Concrete.  Fuller's  Rule  for  Proportioning  Concrete.  Depth  of 
Concrete  Covering  of  Steel. 

CHAPTER  IV 

ANALYSIS  OF  STRESSES     .  37 

BEAM  THEORIES. 

Navier's  Hypothesis.  Hooke's  Law.  The  Straight  Line  Analy- 
sis. Nomenclature.  Excessive  Reinforcement.  Under  Reinforce- 
ment. Double  Reinforcement.  Effect  of  Compressive  Steel. 

T-BEAMS. 

Stresses.  Tests  of  T-beams.  Influence  of  Web  Compression. 
Depth  of  T-beams. 

Flexural  Stresses  under  Ultimate  Loads.  Parabolic  Stress-Strain 
Diagram.  General  Parabolic  Stress-Strain  Relutfon. 

Diagonal  Tension  in  Beams.  Distribution  of  Stresses  in  Homo- 
geneous Beams.  Shear  and  Axial  Stress.  Distribution  of  Stresses 
in  Reinforced  Beams.  Manner  of  Failure  of  Beams.  Working 
Shearing  Stresses.  Stresses  in  Web  Reinforcement.  Spacing  of 
Web  Reinforcement.  Points  of  Bending-up  Bars.  Beams  not 
needing  Web  Reinforcement.  Bond  Stresses  in  Bent-up  Bars. 
Transverse  Spacing  of  Reinforcement.  Working  Rule  for  Trans- 
verse Spacing.  Inclination  of  Bent-up  Rods.  Stresses  in  Concrete 
due  to  Bent-up  Rods.  Water-proof  and  Fire-proof  Qualities  of 
Concrete.  Temperature  Stresses.  -Reinforcement  for  Tempera- 
ture Stresses.  Combined  Flexural  and  Axial  Stresses  in  Beams. 
Case  I,  Rectangular  Section  with  Symmetrical  Reinforcement,  all 
in  Compression.  Case  II,  Tension  and  Compression  in  the  Sec- 
tion. Flexure  and  Axial  Tension. 

COLUMNS  OF  CONCRETE. 

Columns  with  Longitudinal  Reinforcement.  Tests  of  Plain 
Columns.  Tests  of  Columns  with  Longitudinal  Reinforcement. 
Working  Loads  for  Columns  with  Longitudinal  Reinforcement. 
Columns  with  Hooped  Reinforcement.  Stresses  in  Hoops.  Tests 
of  Hooped  Columns.  Discussion  of  Tests  of  Columns.  Strength 
added  by  the  Hoops.  Working  Loads  for  Hooped  Columns. 
Columns  with  Hooped  and  Longitudinal  Reinforcement.  Con- 
sidere's  Tests.  Working  Loads.  Long  Columns. 

BEAMS  SUPPORTED  IN  MORE  THAN  TWO  PLACES. 

Continuous  Beams.  General  Formulas.  Working  Moments 
in  Continuous  Beams.  Slabs  Supported  on  Four  Sides.  Distribu- 


CONTENTS  ix 

PAGE 

tion  of   Floor  Loads  to   Beams.     Girderless  Floors.     Test  of  a 
Girderless  Floor. 

CHAPTER  V 
ELEMENTARY  DESIGN  .    , 132 

ECONOMICAL  PROPORTIONS  OF  BEAMS. 

(a)  When  the  Breadth  is  Constant,  (b)  When  the  Depth  is 
Constant,  (c)  When  the  Area  of  the  Cross-sections  is  Constant. 
(d)  When  the  Ratio  of  Breadth  to  Depth  is  Fixed.  Economic  Pro- 
portions of  T-beams. 

REPORT  OF  THE  JOINT  COMMITTEE. 

Preparation  and  Placing  of  Mortar  and  Cement.  Details  of 
Construction.  Design.  Working  Stresses. 

DESIGN  OF  A  T-BEAM. 

Cross-section  According  to  Shear.  Flange  Width  According  to 
Moments.  The  Reinforcement.  Points  for  Bending-up  Bars. 
Spacing  of  Web  Reinforcement.  The  Stirrups-spacing.  Inves- 
tigation of  the  Flange.  Diagram  for  Solution  of  Slabs  and  Beams. 
Sequence  of  Computations  in  the  Design  of  T-beams. 

DESIGN  OF  REINFORCED  CONCRETE  RETAINING  WALLS. 

The  Lateral  Pressure  of  Earth.  Equivalent  Fluid  Pressure. 
Types  of  Retaining  Walls. 

DESIGN  OF  A  CANTILEVER  RETAINING  WALL. 

The  Vertical  Wall.  Reinforcement  in  Vertical  Wall.  Shear  at 
Top  of  Footing.  Design  of  the  Footing.  Position  of  the  Vertical 
Wall  on  the  Slab.  The  Bearing  Power  of  the  Soil.  The  Outer 
Cantilever.  The  Inner  Cantilever.  Anchoring  the  Reinforcement. 
Drainage.  Sequence  of  Work  in  the  Design  of  Cantilever  Retain- 
ing Walls. 

DESIGN  OF  A  COUNTERFORT  RETAINING  WALL. 

The  Vertical  Slab.  The  Footing.  The  Outer  Cantilever.  The 
Inner  Floor  Slab.  The  Counterfort.  The  Reinforcement.  Se- 
quence of  Work  in  the  Design  of  a  Counterfort  Retaining  Wall. 

THE  ARCH.    ALGEBRAIC  ANALYSIS. 

Types.  The  Three  Hinged  Arch.  The  Two  Hinged  Arch.  The 
Hingeless  Arch.  Stresses  at  any  Section.  Stresses  due  to  Tem- 
perature. Effect  of  Arch  Ring  Shortening  due  to  Thrust. 

THE  ARCH.     GRAPHIC  ANALYSIS. 

Principles.  Conditions  of  Equilibrium.  Location  of  k'  k'.  The 
Trial  Equilibrium  Polygon.  The  True  Equilibrium  Polygon. 
Temperature  Stresses.  Stresses  due  to  Arch  Ring  Shortening. 


x  CONTENTS 

PAGE 

DESIGN  OF  A  REINFORCED  CONCRETE  ELASTIC  ARCH  RING. 

Loads  and  Dimensions.  Depth  of  the  Keystone.  Curves  of 
the  Intrados  and  Extrados.  Making  d  s  -j-  /  constant.  The  Re- 
duced Load  Contour.  The  Load  Diagram.  The  Trial  Equilibrium 
Polygon.  The  True  Equilibrium  Polygon.  Stresses  due  to  Loads. 
Temperature  Stresses.  Stresses  due  to  Arch  Ring  Shortening. 
Final  Tension  and  Compression  Stresses.  Shearing  Stresses. 
Final  Dimensions.  Abutments.  Sequence  of  Work  in  Arch  De- 
sign. 

DESIGN  OF  A  BEAM  AND  GIRDER  FLOOR. 

The  Beams.     The  Girders.     The  Slabs. 
DESIGN  OF  A  REINFORCED  CONCRETE  HOLLOW  DAM. 

Sequence  of  Work  in  Computation. 
INDEX  217 


REINFORCED  CONCRETE 


CHAPTER  I 

HISTORICAL  SKETCH 

THE  device  of  adding  material  having  tensile  strength  to 
those  that  are  deficient  in  this  respect  has  long  been  practiced 
by  the  building  trades.  The  order  of  the  Egyptian  taskmasters 
to  their  servants  that  they  make  "  brick  without  straw,"  had 
reference  to  the  custom  of  mixing  some  such  material  to  clay 
to  increase  its  strength.  The  same  practice  is  common  in  Cen- 
tral America,  where  straw  is  mixed  with  mud  to  reinforce  the 
adobe  or  sun  dried  brick.  It  is  everywhere  common  to  incor- 
porate animal  hair  with  lime  mortar  to  make  the  plaster  for 
interior  walls.  Centuries  ago  boats  in  common  use  on  the 
Nile  were  of  wicker  frame  with  a  covering  of  clay.  The  clay 
was  deficient  in  tensile  strength,  and  the  wood  served  as  rein- 
forcement. Similar  boats,  having  iron  instead  of  wood  and 
concrete  instead  of  clay,  are  in  rather  common  use  to-day  in 
Italy  and  also  in  Panama.  The  roof  of  a  tomb  on  Via  Appia, 
Rome,  is  said  to  have  been  found  to  consist  of  cement  with 
embedded  bronze  rods.  These  tombs  were  built  about  100  B.  c. 
In  1830,  in  a  book  called  the  "  Encyclopaedia  of  Cottage,  Farm 
and  Village  Architecture,"  it  was  suggested  that  roofs  might  be 
constructed  of  iron  rods  thickly  embedded  in  cement.  In  1840 
floors  were  made  of  plaster  of  Paris  and  iron  rods  in  France, 
and  in  1854  a  mason  named  W.  B.  Wilkinson  is  said  to  have 
taken  out  a  patent  for  a  floor  of  concrete  and  iron  in  New- 
castle-on-Tyne.  Even  modern  methods  of  construction  seem 
to  be  old.  Recent  discoveries  in  Central  America  show  ancient 
buildings  that  have  apparently  been  poured  as  a  unit  between 
wooden  forms. 

While  some  of  the  accounts  of  the  above  named  practices  are 
somewhat  in  the  nature  of  rumors,  it  is  certain  that  what  is 

1 


REINFORCED   CONCRETE 


now  commonly  known  as  reinforced  concrete  is  but  a  natural 
adaptation  of  very  old  practices  to  modern  building  material. 
From  straw  and  reeds  or  hair  with  mud  or  lime  it  was  but  a 
step  to  the  use  of  steel  and  concrete.  The  use  of  more  expensive 
material  naturally  prompted  the  study  of  methods  to  determine 
the  most  advantageous  distribution  of  the  metal. 

The  credit  for  demonstrating  the  advantages  of  the  combina- 
tion of  iron  and  concrete  in  engineering  structures  seems  to 
belong  to  a  Parisian  gardener  named  Jean  Monier.  Fortunately 
he  was  an  engineer  by  instinct,  if  not  by  training,  and  having 
interested  capitalists  in  his  schemes,  he  began  a  highly  suc- 
cessful career  as  a  builder  of  water 
and  gas  tanks,  sewers,  arches,  and 
floors.  In  a  paper  read  in  1894  be- 
fore the  American  Society  of  Civil 
Engineers,  Fr.  von  Emberger,  C.  E., 
member  of  the  Austrian  Society  of 
Engineers  and  Architects,  stated 
that  this  method  of  engineering  con- 
struction was  then  sixteen  years  old 
and  that  Monier  began  his  work  of 
this  character  in  1875.  The  discus- 
sion of  this  paper  brought  out  the 
fact  that  European  engineers  were, 
even  then,  building  reinforced  con- 
crete arches  of  over  150  feet  span,  and  of  only  six  inches  to 
eight  inches  depth  at  the  crown. 

In  this  country  the  development  of  reinforced  concrete  con- 
struction was  hardly  less  rapid.  In  "  Transactions  of  the  Amer- 
ican Society  of  Mechanical  Engineers,"  Vol.  IV,  1888,  Mr.  W.  E. 
Ward  tells  of  the  use  of  light  iron  rods  and  beams  embedded 
in  the  concrete  walls,  floors,  and  cornices  of  a  building  erected 
by  him  in  Port  Chester,  N.  Y.,  in  1875.  The  construction 
shown  in  Fig.  1  is  described  in  Engineering  News,  Sept.  8,  1888, 
in  its  digest  of  a  paper  read  before  the  Technical  Society  of 
the  Pacific  Coast,  by  Mr.  G.  W.  Percy.  The  lintel  shown  had 
a  span  of  15  feet,  a  depth  of  2  feet  10  inches,  and  carried  three 
stories  of  brick  wall.  What  is,  possibly,  the  first  reinforced 
concrete  bridge  in  this  country  w"as  built  with  a  span  of  35  feet 
in  the  Golden  Gate  Park,  San  Francisco,  in  1889. 


FIG.  1. 


HISTORICAL   SKETCH 


Since  the  dates  mentioned  above,  this  style  of  construction 
has  steadily  found  increasing  favor  among  engineers  and  archi- 
tects, and  there  is  scarcely  a  type  of  structure  that  is  not  some- 
times made  in  this  manner.  Reinforced  concrete  may  be  used 
in  nearly  all  cases  where  stone  would  be  acceptable,  and  in  many 
other  cases  where  the  latter  is  unsuitable  on  account  of  the 
lack  of  tensile  strength. 

While  there  is  no  definite  information  at  hand  as  to  the  amount 
of  reinforced  concrete  work  done  yearly  in  this  country,  some 
indication  of  the  facts  may  be  seen  by  an  inspection  of  the  statis- 
tics regarding  the  production  of  Portland  cement  in  recent  years. 


Year 

Production 

Imports 

Year 

Production 

Imports 

1880 

42,000 

187,000 

1898 

3,584,586 

2,119,880 

1885 

150,000 

— 

1900 

8,482,000 

— 

1888 

— 

1,843,814 

1903 

22,342,973 

— 

1890 

335,500 

— 

1905 

35,246,812 

— 

1891 

454,813 

— 

1908 

51,002,612 

839,247 

1895 

990,324 

— 

1909 

62,508,461 

— 

The  table  indicates  the  production  in  and  imports  into  United 
States,  in  barrels.  The  marked  increase  in  production  after 
1900  was,  in  part,  due  to  the  introduction  of  the  rotary  kilns 
and  the  use  of  coal  as  fuel.  At  the  same  time  the  importation 
was  at  its  height,  and  it  has  since  dwindled  to  an  almost  negligible 
quantity.  Previous  to  1890  most  of  the  cement  made  in  this 
country  was  of  the  " natural"  variety,  which,  though  well 
adapted  to  certain  uses,  is  seldom  used  in  reinforced  construc- 
tion. The  increase  in  the  production  of  Portland  cement  and 
the  corresponding  decrease  in  that  of  the  natural  is  shown  for 
three  years,  in  barrels,  in  the  following  table: 


Variety 

1907 

1908 

1909 

Natural 

2,887,700 

1,686  682 

1  527  279 

Portland 

48,785,390 

51,072,612 

62  508  461 

Of  course,  not  all  of  this  decided  increase  in  the  production 
of  cement  can  be  attributed  to  the  growth  of  the  reinforced 


4  REINFORCED    CONCRETE 

construction,  but  there  is  undoubtedly  some  relation  between 
the  two  branches  of  industry. 

The  enduring  qualities  of  steel-concrete  construction  have 
not  yet,  of  course,  been  fully  proved;  but  the  severe  tests  through 
which  such  buildings  passed  at  the  times  of  the  Baltimore  fire 
and  the  San  Francisco  disaster  have  gone  far  toward  removing 
any  doubts  some  may  have  had  at  first.  Reinforced  concrete 
construction  of  many  kinds  is  being  done  as  cheaply  as  of  steel, 
and,  in  numerous  instances,  its  ability  to  resist  heat,  dampness, 
smoke,  and  other  gases  makes  it  preferable  to  any  other  building 
material.  There  is  little  doubt  that  iron  embedded  in  concrete 
will  resist  corrosion  indefinitely,  as  has  been  shown  in  several 
instances  where  anchor  bars  have  been  removed  after  more 
than  a  century  of  use  and  exposure. 

Reinforced  concrete  is  found  in  the  deepest  foundations,  and 
even  in  the  piles  upon  which  the  masonry  rests.  It  forms  the 
roof  of  the  structure,  and  the  spires  and  highest  chimneys  are 
made  of  this  material.  It  is  in  use  in  the  floors,  the  ceilings, 
the  walls,  the  stairs,  the  cornices,  and  the  handrailings  of  build- 
ings of  every  kind.  The  railroad  engineer  uses  reinforced  con- 
crete for  ties,  for  culverts,  for  arches,  for  tunnel  portals,  for 
round  houses,  bumping  posts,  and  for  trainsheds.  The  hydraulic 
engineer  builds  his  dam  in  the  wilderness  and  his  fountains  in 
the  city  of  a  combination  of  cement  and  iron.  The  surveyor 
uses  reinforced  concrete  boundary  fence  posts,  and  the  electrical 
engineer  uses  poles  for  transmission  lines  similarly  strengthened. 
The  municipal  engineer  in  his  designs  of  sidewalks,  curbing, 
culverts,  and  sewer  pipe,  and  the  naval  architect  in  his  rather 
experimental  designs  of  barges,  testify  to  the  great  variety  of 
uses  to  which  this  type  of  construction  may  be  adapted. 

The  remarkable  development  of  this  new  building  construc- 
tion has  not  been  accomplished  without  many  serious  and  re- 
grettable failures.  Many  lives  and  much  property  have  been 
destroyed  in  collapses  of  buildings  erected  according  to  improper 
designs  and  lack  of  intelligent  inspection  during  the  building. 
Men  without  engineering  training,  who  would  not  attempt  the 
simplest  design  in  structural  steel,  have  undertaken  the  design 
and  erection  of  structures  really  demanding  the  services  of  the 
most  expert.  Reinforced  concrete  was  exploited  as  a  building 
material  that,  in  point  of  cost,  would  revolutionize  engineering 


HISTORICAL   SKETCH  5 

construction,  and  extravagant  claims  were  made  as  to  its  cheap- 
ness. In  the  attempt  to  substantiate  these  claims,  practices 
which  tended  to  bring  this  class  of  construction  into  disrepute 
became  common.  However,  the  industry  has  passed  its  period 
of  trial,  and  it  is  recognized  that,  when  properly  designed  and 
erected,  safe  and  economical  results  may  be  effected  with  its 
use.  The  claim  of  economy  is  by  no  means  abandoned,  but  it 
is  based  upon  the  ultimate  rather  than  upon  the  initial  cost. 


CHAPTER   II 
THE  COMPONENT  PARTS 

Cement.  It  is  assumed  that  the  reader  is  familiar  with  the 
composition,  appearance,  and  usual  methods  of  testing  Port- 
land cement,  and  it  is  not  necessary  to  review  such  information 
here.  The  following  specifications  have  been  approved  by  the 
American  Society  for  Testing  Materials,  the  American  Society 
of  Civil  Engineers,  and  the  American  Railway  Engineering  and 
Maintenance  of  Way  Association. 

PORTLAND    CEMENT1 

Definition.  This  term  is  applied  to  the  finely  pulverized  product 
resulting  from  the  calcination  to  incipient  fusion  of  an  intimate  mixture 
of  properly  proportioned  argillaceous  and  calcareous  materials,  and  to 
which  no  addition  greater  than  3  per  cent,  has  been  made  subsequent 
to  calcination. 

Specific  Gravity.  The  specific  gravity  of  the  cement,  ignited  at  a 
low  red  heat,  shall  not  be  less  than  3.10;  and  the  cement  shall  not  show 
a  loss  on  ignition  of  over  4  per  cent. 

Fineness.  It  shall  have  by  weight  a  residue  of  not  more  than  8  per 
cent  on  the  No.  100,  and  not  more  than  25  per  cent,  on  the  No.  200  sieve. 

Time  of  Setting.  It  shall  not  develop  initial  set  in  less  than  thirty 
minutes,  and  must  develop  hard  set  in  not  less  than  one  hour  nor  more 
than  ten  hours. 

Tensile  Strength.  The  minimum  requirements  for  tensile  strength 
for  briquettes  one  inch  square  in  section  shall  be  within  the  following 
limits,  and  shall  show  no  retrogression  in  strength  within  the  periods 
specified: 

Neat  Cement 
Age  Strength 

24  hours  in  moist  air 150  -  200  Ib. 

7  days  (1  day  in  moist  air,  6  days  in  water) 450  -  550  " 

28  days  (1  day  in  moist  air,  27  days  in  water) 550  —  650  " 

1  Natural  cement  is  not  considered  here  ao  it  is  seldom  used  in  reinforced 
concrete  construction. 

6 


THE    COMPONENT    PARTS  7 

One  Part  Cement,  Three  Parts  Sand1 

7  days  (1  day  in  moist  air,  6  days  in  water) 150  —  200  Ib. 

28  days  (1  day  in  moist  air,  27  days  in  water) 200  -  300  " 

Constancy  of  Volume.  Pats  of  neat  cement  about  three  inches  in 
diameter,  one-half  inch  thick  at  the  center,  and  tapering  to  a  thin  edge, 
shall  be  kept  in  moist  air  for  a  period  of  twenty-four  hours. 

(a)  A  pat  is  then  kept  in  air  at  normal  temperature  and  observed  at 
intervals  for  at  least  28  days. 

(6)  Another  pat  is  kept  in  water  maintained  as  near  70°  F.  as  prac- 
ticable, and  observed  at  intervals  for  at  least  28  days. 

(c)  A  third  pat  is  exposed  in  any  convenient  way  in  an  atmosphere 
of  steam,  above  boiling  water,  in  a  loosely  closed  vessel  for  five  hours. 

These  pats,  to  satisfactorily  pass  the  requirements,  shall  remain  firm 
and  hard  and  show  no  signs  of  distortion,  checking,  cracking,  or  disin- 
tegrating. 

Sulphuric  Acid  and  Magnesia.  The  cement  shall  not  contain  more 
than  1.75  per  cent,  of  anhydrous  sulphuric  acid  (S02),  nor  more  than 
4  per  cent,  of  magnesia  (MgO). 

Sand 

Specifications  for  sand  are  not  nearly  as  definite  or  as  complete  as 
those  for  cement,  and  the  two  that  are  given  here  are  those  in  common 
use.  They  are  both  recommended  by  the  American  Railway  Engineer- 
ing and  Maintenance  of  Way  Association,  the  latter  specification  having 
special  reference  to  reinforced  concrete. 

Sand.  The  sand  shall  be  clean,  sharp,  coarse,  and  of  grains  varying 
in  size.  It  shall  be  free  from  sticks  and  other  foreign  matter,  but  it 
may  contain  clay  or  loam  not  to  exceed  five  per  cent.  Crusher  dust, 
screened  to  reject  all  particles  over  one-quarter  inch  in  diameter,  may  be 
used  instead  of  sand  if  approved  by  the  engineer. 

Fine  aggregate  shall  consist  of  sand,  crushed  stone,  or  gravel  screenings, 
graded  from  fine  to  coarse,  and  passing,  when  dry,  a  screen  having  one- 
quarter-inch  diameter  holes;  it  shall  preferably  be  of  siliceous  material, 
clean,  coarse,  free  from  vegetable  loam  or  other  deleterious  matter,  and 
not  more  than  six  per  cent,  shall  pass  a  sieve  having  100  meshes  per  linear 
inch. 

Some  of  the  most  thorough  experiments  upon  the  effect  of 
the  quality  of  sand  upon  mortar  have  been  made  by  the 
United  States  Geological  Survey  and  published  as  Bulletin. 

1  Required  by  Am.  Ry.  Engineering  and  Maintenance  of  Way  Assn. 
Mortars  made  of  one  part  Portland  cement  and  three  parts  fine  aggregate  by 
weight  when  made  into  briquettes  shall  show  a  tensile  strength  of  at  least 
70  per  cent,  of  the  strength  of  1  to  3' mortar  of  the  same  consistency  made  with 
the  same  cement  and  standard  Ottawa  sand. 


(C15) 


(BU) 


(A  13) 


FIG.  2 


THE    COMPONENT   PARTS 


9 


No.  331.  Like  most  investigations  the  results  are  somewhat  con- 
flicting, but  seem  to  indicate  that  a  sieve  analysis  may  be  relied 
upon  to  a  considerable  extent  as  a  test  of  fitness  of  a  sand. 
Photographs  of  three  of  the  twenty-two  samples  tested  are 
reproduced  in  Fig.  2,  and  sieve  analysis  diagrams  of  the  same 
are  taken  from  the  above-named  bulletin. 

The  photograph  at  the  top  shows  the  appearance  of  the  poorest 
of  the  sands  tested,  that  in  the  middle  one  of  about  average 

Diameters  of  ParticlesTin  Inclies 

<M  CO  -*•  WJ  «p  »-«  OO 

o  o  o  o  o  o  o 


1UU 

80 
60 
40 
20 

I 

>> 

^-* 

--  •- 

£' 

.—  "• 

" 

/r 

/ 

7 

______ 

- 

— 

,—  — 

— 

--—  — 

/ 

/ 

___  . 

,  — 

— 

— 

•— 

" 

|f  8   §  3       8            £                                            S 
Numbers  of  Sieves 

FIG.  3 

grade,  and  that  at  the  bottom  of  the  page  one  of  the  very  best. 
This  rating  is  on  the  basis  of  compressive  strength  at  the  end 
of  28  days.  The  even  distribution  of  the  large  and  small  grains 
is  at  once  apparent  in  A,  and  the  lack  of  large  grains  is  as  clearly 
shown  in  C.  If  no  other  information  were  at  hand,  B  would 
naturally  be  classed  as  about  an  average  of  the  other  two.  The 
compressive  strengths  of  these  samples  in  1  to  3  mixtures  at  the 
end  of  28  days  were:  A,  5582;  B,  3159;  and  C,  1898  pounds 


10 


REINFORCED    CONCRETE 


per  square  inch.  When  samples  of  B  were  tested  in  the  same 
way,  with  all  grains  omitted  except  those  passing  a  No.  30 
sieve  and  held  on  a  No.  40  sieve,  the  strength  was  reduced  to 
1969  pounds  per  square  inch. 

The  diagrams,  Fig.  3,  show,  by  the  full  line  curve,  the  pro- 
portions passing  each  sieve  as  indicated  on  the  lower  line;  the 
straight  dotted  line  is  that  of  a  uniform  gradation  from  zero 
to  100  per  cent,  on  the  ordinate  for  0.25  inch  diameter,  since  all 
sand  was  such  as  to  pass  that  mesh.  Then  the  area  between 
this  line  and  that  of  the  sieve  analysis  curve  indicates  the 
departure  of  the  sample  from  the  uniform  gradation. 

The  weights  per  cubic  foot  of  the  various  sands  ranged  from 
89  pounds  to  119.9  pounds;  the  per  cent,  of  voids  from  26.9 
to  40.9,  and  the  density  of  the  mortar  from  0.676  to  0.808.  If 
the  compressive  strength  of  the  1  to  3  mortar  at  the  end  of 
28  days  be  the  criterion  of  value,  the  sands  are  ranked  according 
to  the  second  column  of  the  following  table.  The  figures  in 
the  other  columns  indicate  the  rating  according  to  headings. 


No. 

28  Days 

180  Days 

Voids 

Sieve 

Density 

Weight 

16 

1 

3 

5 

1 

7 

2 

19 

2 

4 

1 

5 

3 

1 

13 

3 

2 

4 

3 

9 

2 

8 

4 

1 

6 

4 

6 

7 

20 

5 

5 

2 

8 

2 

5 

4 

6 

6 

2 

2 

1 

4 

22 

7 

10 

18 

12 

17 

18 

6 

8 

14 

8 

11 

5 

8 

3 

9 

9 

17 

7 

10 

17 

9 

10 

18 

7 

6 

4 

10 

2 

11 

8 

16 

10 

8 

13 

7 

12 

7 

8 

13 

16 

6 

14 

13 

15 

11 

14 

15 

9 

17 

14 

12 

14 

18 

18 

15 

18 

15 

17 

13 

15 

14 

14 

10 

16 

13 

8 

17 

12 

11 

5 

17 

11 

15 

9 

11 

16 

21 

18 

19 

20 

19 

19 

20 

1 

19 

16 

12 

16 

13 

12 

15 

20 

20 

19 

20 

20 

19 

Two  of  the  sands,  Nos.  11  and  12,  were  omitted  from  the 


THE   COMPONENT    PARTS  11 

above  comparison  since  they  were  mixtures  of  sand  and  stone 
screenings. 

It  should  be  noted  that  the  ranking  by  weight,  by  voids,  by 
sifting,  or  by  compressive  strength  of  the  best  six  gives  nearly 
identical  results.  The  poorer  samples,  also,  are  detected  by 
any  of  these  methods  with  nearly  the  same  certainty.  The 
density  of  the  mortar  does  not  seem  to  indicate  the  strength 
with  any  greater  accuracy  than  does  the  weight  of  the  sand. 

From  information,  obtained  otherwise,  concerning  mortar 
made  from  various  sands  in  general  use  in  Illinois  it  is  reason- 
ably certain  that  the  strength  of  mortar  from  the  best  sands  is 
to  that  made  from  the  poorest  about  as  450  is  to  150.  Also, 
that  1  to  3  mortar  from  the  best  is  stronger  than  1  to  1  mortar 
from  the  poorest.  In  making  these  tests  the  same  percentage 
of  water  was  used  with  the  fine  as  with  the  coarse  sand,  and 
so  the  results  are  not  entirely  conclusive.  As  the  results  are 
based  on  the  strength  of  mortar  at  the  end  of  90  days,  however, 
there  should  be  little  or  no  error  in  the  inferences. 

It  would  seem  from  the  foregoing  that  it  is  economy  to  use 
the  better  grades  of  sand,  even  at  great  expense  for  transporta- 
tion. When  the  prices  of  cement  and  of  the  several  varieties 
of  available  sands  are  known,  the  relative  economy  of  various 
mixtures  may  be  computed. 

In  the  tests  of  the  United  States  Geological  Survey,  noted 
above,  were  included  those  on  gravel  and  broken  stone  screen- 
ings. The  results  showed  that  either  of  these  may  be  safely 
used  instead  of  sand,  and  in  general  with  even  better  results. 

By  means  of  three  tests  that  are  readily  made  the  relative 
value  of  various  sands  may  be  judged  quite  accurately.  These 
tests  are: 

(a)  The  appearance. 
(6)  The  feeling, 
(c)  The  weight. 

The  better  sands  show  a  generous  sprinkling  of  coarse  grains 
mixed  with  the  fine  material  and  intermediate  gradations. 

The  grains  should  be  of  irregular  shapes  even  though  smooth; 
but  sharpness  is  desirable.  Upon  rubbing  the  sand  in  the  palm 
of  the  hand,  traces  of  clay  should  be  seen. 

The  heavier  the  sand  the  better.  Well  shaken  sand  should 
weigh  over  100  pounds  per  cubic  foot  when  dry. 


12  REINFORCED    CONCRETE 

The  Aggregate 

The  Aggregate.  Coarse  aggregate  shall  consist  of  crushed  stone  or 
gravel,  graded  in  size,  and  which  is  retained  on  a  screen  having  one- 
quarter  inch  diameter  holes;  it  shall  be  clean,  hard,  durable,  and  free 
from  all  deleterious  material.  Aggregates  containing  soft,  flat,  or  elon- 
gated particles  shall  not  be  used. 

The  maximum  size  of  the  coarse  aggregate  shall  be  such  that  it  will 
not  separate  from  the  mortar  in  laying  and  will  not  prevent  the  concrete 
fully  surrounding  the  reinforcement  or  filling  all  parts  of  the  forms. 
Where  concrete  is  used  in  mass  the  maximum  size  of  the  coarse  aggregate 
may,  at  the  option  of  the  engineer,  be  such  as  to  pass  a  3-inch  ring.  For 
reinforced  concrete,  sizes  are  usually  not  to  exceed  one  inch  in  any  direc- 
tion, but  may  be  varied  to  suit  the  character  of  the  reinforcement. 

The  Water.  The  water  used  in  mixing  concrete  shall  be  free  from 
oil,  acid,  alkalies,  or  vegetable  matter. 

The  Steel.  The  metal  reinforcement  steel  shall  ...  be  free  from 
dust,  scale,  or  coatings  of  any  character  which  would  tend  to  reduce  or 
destroy  the  bond. 

The  ultimate  tensile  strength,  in  pounds  per  square  inch,  shall  be 
60,000  for  structural  steel  and  85,000  for  high  carbon  steel. 

Bending  tests  may  be  made  by  pressure  or  by  blows.  Shapes  and  bars 
less  than  one  inch  thick  shall  bend  through  180°  flat  without  breaking 
it  of  structural,  and  through  180°  around  a  diameter  of  four  times  the 
thickness  if  of  high  carbon  steel. 

Structural  steel  1  inch  thick  and  over,  tested  as  rolled,  shall  bend  cold 
180°  around  a  pin,  the  diameter  of  which  is  equal  to  twice  the  thickness 
of  the  bar,  without  fracture  on  the  outside  of  the  bend. 

Further  specifications  regarding  the  fabrication  of  reinforced 
concrete  will  be  given  in  another  chapter. 


PROBLEM1 

Let  the  mixture  be  1:3:6,  and  let  a  cubic  yard  be  composed  of  0.96 
barrel  of  cement,  0.41  cubic  yards  of  sand,  and  0.82  cubic  yards  of  stone. 
Let  the  cost  of  the  cement  be  $1.80  per  barrel,  and  that  of  the  stone 
$1.25  per  cubic  yard.  Let  sand  at  50  cents  and  at  $1.50  per  cubic  yard 
be  available.  Which  sand  is  the  more  economical  if  the  compressive 
strength  of  the  corresponding  concrete  be  1500  and  1800  pounds  per 
square  inch  respectively?  How  will  the  relation  be  changed  if  the  cost 
of  labor  be  added? 


CHAPTER   III 

TESTS  OF  REINFORCED  CONCRETE 

Sources  of  Information.  Much  of  the  information  regarding 
the  strength  of  building  material  is  derived  from  experiments 
conducted  in  the  testing  laboratories  of  the  various  technical 
schools  of  the  country.  The  results  of  these  tests,  with  con- 
clusions drawn  therefrom  by  the  directors  in  charge,  are  published 
in  pamphlet  form  from  time  to  time,  and  are  for  distribution  at 
nominal  charge.  Some  of  these  experiments  are  conducted  by 
a  regularly  employed  corps  of  men,  some  personally  by  the 
director,  with  occasional  help,  and  some  as  thesis  work  of  senior 
engineering  students  under  careful  supervision. 

The  United  States  Government  maintains  several  labora- 
tories, notably  one  at  Watertown,  Mass.,  where  some  of  the 
most  reliable  information  has  been  gathered.  The  results  of 
this  work  are  published  annually  as  Tests  of  Metals,  a  pamphlet 
that  is  for  distribution  at  moderate  price. 

The  transactions  of  the  national  engineering  societies  contain 
papers  by  eminent  men  concerning  researches  of  this  character. 
The  societies,  as  bodies,  do  not  conduct  tests,  but  the  discus- 
sions bring  out  much  of  the  combined  knowledge  of  the  indi- 
vidual members.  Among  such  societies  may  be  named  the 
American  Society  for  Testing  Materials,  the  Western  Society 
of  Engineers,  the  American  Society  of  Civil  Engineers,  and 
the  American  Railway  Engineering  and  Maintenance  of  Way 
Association. 

Several  of  the  more  local  engineering  organizations  are  united 
in  publishing  the  "  Journal  of  the  Association  of  Engineering 
Societies,"  wdth  headquarters  at  present  in  Boston. 

In  foreign  countries  there  is  as  much  activity  in  collecting 
information  of  this  character  and  in  making  it  known  as  is  the 
case  here.  Everywhere  the  designer  is  restricted  less  by  lack 
of  reliable  tests  than  by  intelligent  interpretation  of  the  same. 

13 


14 


REINFORCED    CONCRETE 


It  is  not  the  purpose  here  to  quote  fully  from  any  of  the 
above  named  sources,  and  what  immediately  follows  is  general 
rather  than  in  detail.  The  values  representing  the  strength 
of  the  materials  under  various  circumstances  of  strain  in  the 
examples  and  problems  in  later  parts  of  the  book  are  supposed 
to  represent  reasonable  practice  only  and  not  definite  recommen- 
dations. 

The  tests  of  reinforced  concrete  are  naturally  divided  into 
two  parts:  those  of  the  component  materials,  as  the  cement, 
mortar,  concrete,  and  steel;  and  those  of  the  combination  of 
the  whole  into  shapes  suitable  for  building  purposes.  Only 
the  latter  are  of  such  individuality  as  to  require  more  than 
casual  reference  here. 

TENSILE  STRENGTH  OF  CONCRETE 


Mixture 

Age 
Days 

Strength  in  Lb.  Per 
Square  Inch. 

Authority 

1:3:6 

50 

178 

1:3:6 

60 

160 

1:3:6 

84 

170 

a 

1:3:0 

87 

278 

1:3 

90 

179 

1:4 

90 

130 

b 

1:3 
1:2:4 

2  years 

28 

220 
142 

1:2:4 

28 

160 

c 

1:2:5 

28 

237 

1:2:5 

90 

359 

1:5 

28 

253 

d 

1:5 

90 

290 

(a)  Bulletin  No.  1  University  of  Illinois  Engineering  Experiment  Station, 
1904.  Stress  was  applied  to  cylinders  8"  diameter  and  32"  long,  by  means  of 
a  rod  ending  in  a  socket  joint  in  a  steel  disc,  8"  in  diameter,  bolted  to  each 
end  of  the  specimen.  Unless  otherwise  stated  the  author  of  all  bulletins  of 
the  University  of  Illinois  Engineering  Experiment  Station,  from  which  quota- 
tions are  made  in  this  book,  is  Professor  A.  N.  Talbot. 

(6)  Eisenbetonbau,  Emil  Moersch,  Zurich.  The  mixture  was  of  cement, 
sand,  and  gravel,  the  exact  proportions  not  being  stated. 

(c)  Bulletin  No.  2,  Vol.  4,  University  of  Wisconsin,  1908. 

(d)  Journal  of  the  Association  of  Engineering  Societies,  September,  1900. 

Tensile  Tests.  In  order  that  tests  of  concrete  in  tension 
may  indicate  what  may  be  expected  as  to  the  strength  of  such 


TESTS    OF    REINFORCED    CONCRETE 


15 


material  in  construction  work,  the  test  pieces  must  be  fairly 
large.  As  the  aggregate  contains  pieces  two  inches  or  more  in 
diameter,  there  are  apt  to  be  many  unfilled  spaces  around  the 
outside  surface,  and  in  small  sections  these  cavities  form  larger 
proportions  of  the  whole  area  than  in  large  ones.  Hence,  results 
obtained  in  breaking  small  pieces  are  not  commensurate  with 
the  strength  of  parts  used  in  actual  construction.  The  diffi- 
culties attending  the  making  of  tensile  tests  are,  in  part,  due 
to  the  fact  that  it  is  very  hard  to  so  plan  the  experiments  as 
to  cause  a  uniform  distribution  of  the  load  over  the  whole  cross 
section  if  its  area  be  large.  As  concrete  is  seldom  required  to 
resist  direct  tension,  the  lack  of  conclusive  tests  of  this  kind  is 
of  little  comparative  consequence. 

The  tests  on  beams  furnish  an  indirect  method  of  finding  the 
tensile  strength  by  computation.  Results  derived  in  this  way 
show  a  modulus  of  rupture  for  plain  concrete  beams  nearly 
double  the  results  given  above.  In  the  bulletin  (a)  above,  the 
following  tests  of  plain  concrete  beams  are  given. 

TENSION  IN  PLAIN  CONCRETE  BEAMS 


Mixture 

Age 
Days 

Section 
Square  Inch 

Span 

Strength 
Lb.  Per  Sq.  In. 

1:3:6 

64 

12  "  X  13|" 

14' 

412 

65 

14' 

337 

64 

14' 

322 

62 

10'  8" 

390 

62 

10'  8" 

355 

61 

8'  6" 

347 

62 

8'  6" 

422 

61 

5' 

299 

64 

5' 

299 

The  principal  use  of  the  tensile  strength  in  beams  is  in  taking 
some  of  the  diagonal  stresses,  as  will  be  shown  later.  The  rea- 
son for  the  discrepancy  between  the  values  in  the  last  table 
and  those  derived  from  direct  tension  will  be  more  fully  ex- 
plained later.  Briefly,  the  reason  is  that,  for  great  and  small 
deformations,  the  corresponding  stresses  are  not  in  the  same 
ratio  either  on  the  tension  or  compression  sides  or  as  between 
the  two. 

The  practice  is  to  disregard  the  tensile  strength  of  concrete 


16 


REINFORCED    CONCRETE 


and  to  assume  that  all  the  tensile  stresses  are  carried  by  the 
steel.  The  safe  tension  combined  with  shear,  or  diagonal  ten- 
sion in  beams,  is  assumed  as  50  pounds  per  square  inch. 

Compression  Tests.  As  concrete  is  used  in  direct  compres- 
sion, and  as  stresses  of  this  kind  occur  during  flexure,  many 
tests  have  been  made  to  determine  the  reliable  unit  strength. 


COMPRESSIVE  STRENGTH  OF  CONCRETE 


Mixture 

Age 
Days 

Strength 
Lb.  Per  Sq. 
Inch 

Authority 

Mixture 

Age 
DayS 

Strength 
Lb.  Per  Sq. 
Inch 

Author- 
ity 

1:H:3 

355 

4590 

1:2:4 

58 

3210 

1:2:4 

330 

'3175 

1:2:4 

60 

2620 

83 

ai 

1:4:8 

343 

1315 

:2:4 

3080 

1:4:8 

202 

1294 

:5:10 

60 

903 

1:5:10 

334 

1157 

a 

:5:10 

64 

1098 

1:5:10 

184 

819 

:5:10 

69 

897 

a 

1:2:4 

267 

2110 

:2:4 

34 

407 

:2:4 

17 

664 

1:1:2 

66 

4238 

ai 

:2:4 

11 

1517 

1:3:6 

60 

2428 

:2:4 

6 

836 

1:3:6 

65 

1400 

:4:8 

62 

1573 

1:5:10 

69 

1115 

:3:6 

180 

3088 

1:2:4 

180 

3826 

:3:6 

90 

2522 

1:2:4 

'   90 

2896 

:3:6 

30 

2164 

b 

1:2:4 

30 

2399 

6 

:3:6 

7 

1311 

1:2:4 

7 

1565 

:2:5 

30 

769 

c 

1:1:3 

30 

1466 

1:2:3 

30 

1098 

c 

:2:5 

30 

750 

1:2:4 

30 

904 

:3:6 

30 

550 

d 

:3:6 

30 

450 

1:2:4 

30 

1000 

d 

(a)  Bulletin  No.  29,  Univ.  of  111.,  1908,  Engineering  Experiment  Station. 
Each  result  as  given  is  the  average  of  several  tests,  there  being  thirty  of  these 
in  the  sixty  day,  1:2:4  result.  In  the  tests  of  specimens  over  six  months 
old  the  age  given  is  the  average  of  several  within  narrow  limits.  The  sand, 
stone,  and  the  several  varieties  of  cement  used  were  considered  to  represent 
first  class  practice.  The  test  pieces  were  6"  cubes,  (ai)  Bulletin  No.  8. 

(6)  Tests  were  made  for  Boston  Elevated  Railway  Company  at  the  Water- 
town  Arsenal.  Tests  of  Metals  1899.  The  test  pieces  were  12  *  cubes. 

(c)  Tests  of  Metals,   1898.     Portland  cement  and  cinder  concrete,   12" 
cubes. 

(d)  Journal  Assn.  Engineering  Societies,  Sept.,  1900.     Portland  cement  and 
unscreened  cinder  concrete.     The  test  pieces  were  3|"  X  3£"  X  11". 


TESTS    OF    REINFORCED    CONCRETE 


17 


FIG. 


An  analysis  of  the  above  table  shows  that  concrete  increases 
decidedly  with  the  richness  and  age  of  the  mixture.  Other 
factors  affecting  the  strength  are  the  amount 
of  water  used,  the  care  taken  in  mixing  and 
in  breaking,  and  the  size  and  quality  of  the 
stone.  Even  when  these  factors  are  sup- 
posed to  be  constant,  results  lack  much  of 
exact  agreement.  For  example,  there  seems 
to  be  no  apparent  reason  why,  in  (a)  above, 
the  1:4:8  mixture  at  62  days  should  be  stronger  than  the  same 
mixture  at  343  days  old,  or  than  the  1:3:6  mixture  at  the  age 
of  65  days.  Any  tests  such  as  these,  gathered  from  most  reliable 
sources,  are  only  indicative  and  not  at  all  conclusive  as  to  what 
may  be  expected  under  slightly  different  conditions.  How- 
ever, the  same  conditions  are  found  in  all  building  material, 
to  some  extent,  and  the  factor  of  safety  must  be  chosen 
accordingly. 

A  well  made  1:2:4  concrete  is  usually  assumed  to  carry  600 
pounds  per  square  inch  in  compression,  while  400  pounds  per 
square  inch  may  be  taken  as  a  safe  load  for  a  1:3:6  concrete. 

Tests  of  Shearing.  The  term  shear  is  used  by  different 
experimenters  and  writers  to  express  very  dissimilar  actions. 
The  word,  of  course,  suggests  a  cutting  by  sharp  edges  acting 
in  opposite  directions  and  close  together.  In  this  case  the 
upper  and  lower  fibers  are  broken  or  cut  first,  the  fibers  adjacent 
to  these  are  then  severed,  and,  in  turn,  the  next  and  so  on.  If 
the  opposing  external  forces  be  not  close  together, 
bending  results,  as  is  shown  in  Fig.  4,  if  the  ends 
of  the  piece  subjected  to  shear  be  held  rigidly, 
or  by  single  flexure  if  they  be  not  so  held.  The 
latter  action  is  present  in  loaded  'beams  and 
failure  so  caused  is  often  called  shear,  whereas  it 
more  nearly  resembles  tension.  True  shearing  is 
seldom  accomplished  without  some  attending 
compression.  When  a  prism  is  broken  by  com- 
pressive  forces,  fracture  usually  takes  place  by 
a  diagonal  rather  than  by  an  axial  displacement, 
that  ig>  one  end  takes  an  offset  and  is  pushed  by 
the  other,  as  is  shown  in  Fig.  5.  This  particular 
method  of  failure  is  especially  noticeable  in  wood,  while  stone 


FIG.  5. 


18 


REINFORCED    CONCRETE 


and  concrete  may  break  on  four  sides  in  succession,  leaving  a 
pyramidal  fracture.  If  the  plane  of  fracture  in  Fig.  5  be  at 
45°  with  the  direction  of  the  forces,  the  compressive  strength 
of  the  material  is  fully  two  times  the  shearing  strength,  as  is 
readily  shown  by  the  resolution  of  the  forces. 

In  the  earlier  tests  to  determine  shearing  strength  the  effect 
of  bending  was  not  considered,  and  the  resulting  values  were 
given  as  being  from  one-sixth  to  one-fifth  of  the  compressive 
strength.  The  tests  given  below  were  made  in  a  different 
manner,  and  the  results  are  much  higher. 


Form  of  Specimen 

Mixture 

No. 
of 

Tests 

Strength,    Lb.  Per  Sq.  In. 

Shear  -s-  Com- 
pression 

Shear 

Compression 

Cube 

Cylinder 

Cube 

Cylinder 

1:3:6 

9 

679 

1230 



.55 

_ 

1:3:6 

7 

729 

1230 



.59 



Plain  plate    

1:3:6 

4 

905 

2428 

1322 

.37 

.68 

1:3:6 

1 

968 

1721 

1160 

.56 

.83 

1:2:4 

5 

1193 

3210 

2430 

.37 

.49 

f 

1:3:6 

17 

796 

1230 



.65 



1:3:6 

5 

879 

1230 



.71 



Recessed  block  ... 

1:3:6 

4 

1141 

2428 

1322 

.47 

.86 

1:3:6 

1 

910 

1721 

1160 

.53 

.79 

1:2:4 

5 

1257 

3210 

2430 

.39 

.52 

1:3:6 

4 

1051 

1230 

_ 

.86 

Reinforced  re- 

1:3:6 

4 

1821 

2428 

1322 

.75 

1.38 

cessed  block  . 

1:3:6 

1 

1555 

1721 

1160 

.90 

1.39 

1:2:4 

5 

2145 

3210 

2430 

.67 

.80 

1:3:6 

4 

1313 

2428 

1322 

.54 

1.00 

Restrained  beam  . 

1:3:6 

1 

1020 

1721 

1160 

.59 

.88 

1:2:4 

6 

1418 

3210 

2430 

.44 

.58 

This  table  is  taken  from  Bulletin  No.  8,  University  of  Illinois 
Engineering  Experiment  Station.  The  test  pieces  had,  with 
six  exceptions,  been  stored  for  60  days  in  damp  sand.  The 
first,  sixth,  and  twelfth  were  stored  in  air,  and  the  second, 
seventh,  and  eighth  were  kept  under  water.  The  plain  concrete 
specimens  were  blocks  13  inches  square  and  3  inches  deep. 
The  recesses  in  the  other  blocks  were  slightly  larger  than  the 
plunger,  of  5|  inches  diameter,  which  was  applied  to  the  plain 
side.  In  the  other  blocks  reinforcing  rods  were  inserted  outside 


TESTS    OF    REINFORCED    CONCRETE 


19 


the  area  to  be  punched  to  prevent  splitting.  The  restrained 
beam  was  4  inches  square  in  section,  and  had  a  clear  span  of 
4|  inches,  while  the  plunger  was  4  inches  square. 

The  following  tests  were  made  in  1905,  at  the  Massachusetts 
Institute  of  Technology,  on  cylindrical  beams  5  inches  in  diam- 
eter, clamped  at  the  ends,  and  having  a  clear  span  of  6  inches. 
The  pressure  was  applied  by  a  cast-iron  plunger  having  a  cylin- 
drical bearing  on  the  beam,  so  that,  as  in  the  beam  mentioned 
above,  the  shearing  took  place  through  two  sections. 


Mixture 

Method  of 
Storing 

Shearing  Strength 
Lb.  Per  Square  Inch 

Crushing 
Strength 

Shear  -H 
Compres- 
sion 

Maximum 

Minimum 

Average 

Lb.  Per 
Square  Inch 

1:2:4 

Air 

1630 

960 

1310 

2070 

.63 

1:2:4 

Water 

2090 

1180 

1650 

2620 

.63 

1:3:5 

Air 

1590 

890 

1240 

1310 

.94 

1:3:5 

Water 

1380 

840 

1120 

1360 

.32 

1:3:6 

Air 

1450 

950 

1180 

950 

1.25 

1:3:6 

Water 

1200 

1030 

1120 

1270 

.88 

In  sections  where  shear,  unaccompanied  by  tension,  is  con- 
sidered, it  is  usual  to  assume  120  pounds  per  square  inch  as  the 
working  strength  for  good  1:2:4  concrete.  The  safe  unit  shear 
in  beams  without  diagonal  reinforcement  is  about  40  pounds 
per  square  inch. 

Tests  of  Elasticity.  Although  concrete  is  not  elastic  to  the 
same  extent  that  steel  is,  yet  it  has  the  tendency,  after  the 
application  and  removal  of  light  loads,  to  return  to  its  original 
shape.  If  a  load  be  applied  just  large  enough  to  produce  a 
small  permanent  set  and  then  removed,  other  smaller  loads 
may  be  repeatedly  applied  and  released  without  creating  much, 
if  any,  further  permanent  set.  To  this  extent  concrete  may 
be  considered  sufficiently  elastic  to  warrant  the  application  of 
the  elastic  theory  to  the  analysis  of  stresses. 

Some  of  the  most  reliable  tests  of  the  elasticity  of  concrete 
have  been  made  at  the  Watertown  Arsenal,  and  the  results  may 
be  found  in  " Tests  of  Metals"  since  1898.  The  following  table 
is  from  that  source,  1899. 


20 


REINFORCED    CONCRETE 


Mixture 

Age 

Coefficient  of  Elasticity  Between  Loads, 
in  Lbs.  Per  Square  Inch,  of 

Compresaive 
Strength 
Lbs.  Per  Sq. 
Inch 

100-600 

100-1000 

1000-2000 

1:2:4 

7  days 

2,593,000 

2,054,000 

1,351,000 

1730 

1:2:4 

1  mo. 

2,662,000 

2,445,000 

1,462,000 

2567 

1:2:4 

3  mos. 

3,671,000 

3,170,000 

2,158,000 

2975 

1:2:4 

6  mos. 

3,646,000 

3,567,000 

2,582,000 

3989 

1:3:6 

7  days 

1,869,000 

1,530,000 

— 

1511 

1:3:6 

1  mo. 

2,438,000 

2,135,000 

1,219,000 

2260 

1:3:6 

3  mos. 

2,976,000 

2,656,000 

1,805,000 

2471 

1:3:6 

6  mos. 

3,068,000 

3,503,000 

1,868,000 

3068 

1:6:12 

1  mo. 

1,376,000 

— 

— 

1146 

1:6:12 

3  mos. 

1,642,000 

1,364,000 

— 

1359 

1:6:12 

6  mos. 

1,820,000 

1,522,000 

— 

1592 

'1:2:5 

1  mo. 

1,040,000 

~~ 

~ 

724 

The  concrete  was  made  of  good  quality  Portland  cement, 
bank  sand,  and  broken  conglomerate  stone. 

The  following  table  is  taken  from  Bulletin  No.  10,  Univ.  of 
Illinois  Engineering  Experiment  Station. 


Mixture 

Age 
Days 

Gauged  Length 
Inches 

Coefficient  of 
Elasticity 
Lbs.  Per  Sq.  Inch 

Maximum  Com- 
pressive  Strength 
Lbs.  Per  Sq.  Inch 

1:2:4 

69 

114 

3,150,000 

1722 

1:2:4 

64 

114 

2,530,000 

2004 

1:2:4 

65 

114 

2,500,000 

1615 

1:2:4 

61 

60 

2,370,000 

1709 

1:2:4 

63 

60 

2,000,000 

1189 

1:2:4 

65 

60 

1,490,000 

1079 

It  will  be  seen  that  the  two  sets  of  tests  agree  fairly  well,  and 
also  that,  in  the  same  specimen,  different  coefficients  exist  under 
different  loads. 

Experiments  made  at  the  same  time  on  reinforced  concrete 
columns  give  practically  the  same  values  for  the  coefficient  of 
elasticity  as  those  given  above  for  plain  concrete. 

The  Stress-Strain  Diagram.  Such  diagrams  are  made  by 
plotting  the  observed  unit  stresses  as  ordinates  and  the  corre- 

1  Cinder  concrete. 


TESTS   OF  REINFORCED  CONCRETE 


21 


Deformation-inches 


spending  deformations  as  abscissas.  This  process  will,  with  con- 
crete, produce  a  curve  in  which  the  abscissas  increase  faster  than 
the  ordinates. 

In  Fig.  6  is  shown  a  typical  curve  of  this  kind  for  concrete. 
The  inclination  of  the  tangent  to  the  curve  at  the  beginning 
represents  the  coefficient  of  elasticity  for  small  loads  when  first 
applied,  and  is  sometimes  referred 
to  as  the  "  initial  coefficient "  or 
modulus.  This  tangent  is  prac- 
tically identical  with  the  curve  ab 
until  the  loads  exceed  500  pounds 
per  square  inch.  If  the  load  be 
released  the  deformation  will  not 
become  zero,  but  a  set  as  ac  will 
occur.  When  the  load  is  again 
applied  the  stress-strain  curve  fol- 
lows cb  closely  for  small  loads,  but 
the  curvature  rapidly  increases  as 
the  loads  become  greater.  A  sec- 
ond release  and  a  third  application 
of  the  loads  fixes  another  loop  in 


FIG.  6. 


the  curve,  and  so  on  till  the  load  equals  the  ultimate  strength. 

The  behavior  of  the  concrete  as  shown  by  the  stress-strain 
diagram,  gives  rise  to  some  confusion  as  to  what  is  the  coefficient 
of  elasticity.  The  tangent  of  the  angle  of  inclination  of  the 
straight  line  through  a  and  d  is  called  the  gross  or  secant  coeffi- 
cient, while  that  through  e  and  d  is  the  net  or  elastic  coefficient. 
In  comparing  coefficients  of  elasticity,  it  is  necessary  to  note 
whether  or  not  the  stresses  were  within  the  same  limits.  If 
the  concrete  represented  in  Fig.  6  contain  reinforcement,  it  is 
evident  that  the  stress  in  the  latter  is  indicated  by  a  d'  rather 
than  by  e  d/  and  here  the  gross  coefficient  is  clearly  the  more 
appropriate.  On  the  other  hand,  the  net  coefficient  ddr  -5-  edf 
more  nearly  represents  the  elastic  properties  under  low  stresses 
after  the  second  release  of  the  loads. 

It  is  quite  evident  that,  in  considering  all  stages  of  loading 
from  zero  to  the  ultimate,  a  constant  coefficient  should  not  be 
used,  and  it  is  common  to  assume  the  stress-strain  curve  to  be 
a  parabola  and  make  the  analyses  accordingly.  The  parabola 
in  Fig.  7  has  its  axis  vertical,  and  the  maximum  ordinate  is  the 


22 


REINFORCED  CONCRETE 


ultimate  strength  of  the  concrete.  The  few  experiments  that 
have  been  made  to  determine  the  coefficient  of  elasticity  of 
concrete  in  tension  seem  to  indicate  that,  for  small  stresses,  it 
is  practically  the  same  as  in  compression,  although  probably 
slightly  less. 

In  practice  it  is  usual  to  assume  that  the  ratio  of  the  coeffi- 
cient of  elasticity  of  steel  to  that  of  rich,  well-made  concrete  is 
15,  and  18  to  that  of  the  leaner  mixtures. 

Tests  of  Bond.     The  whole  success  of  reinforced^  concrete 
construction  depends  upon  the  adhesion,  or  bond,  between  the 
metal  and  the  concrete.     This  bond  is  effective  from  two  causes, 
which    are    similar    to    those    holding    a 
wooden   dowel   in   place  in  a  hole  of  a 
/\  diameter  slightly  smaller  than  itself.     In 

the  first  place,  friction,  caused  by  pres- 
sure between  the  surfaces  in  contact, 
makes  withdrawal  difficult.  In  the  sec- 
ond place,  the  dowel  may  be  fixed,  with 
little  friction,  if  the  surfaces  in  contact  be 
covered  with  glue.  Experiments  prove 
pIG  7  that  the  metal  reinforcement  may  be 

moved  slightly  with  reference  to  the  con- 
crete, as  by  a  sharp  blow,  and  still  leave  about  half  of  the  bond 
strength  effective. 

Tests  to  determine  the  strength  of  bond  are  usually  made  by 
pulling  out  a  rod  that  has  been  imbedded  in  a  block  of  concrete. 
A  few  have  been  made  by  forcing  the  steel  out  by  compression. 
In  either  case  the  concrete  is  in  compression,  which  fact  has  the 
tendency  to  increase  the  friction  by  adding  to  the  pressure 
at  the  surfaces  in  contact.  This  condition  is  realized  when  the 
upper  part  of  a  simple  beam  has  reinforcement.  The  steel  in 
the  tension  side  of  a  beam  acts  under  different  conditions.  Here 
the  concrete  is  in  tension  and  often  cracked,  and  the  pressure 
between  the  surfaces  must  become  less.  In  comparing  values 
of  bond  strength  from  different  experiments,  the  manner  of 
making  the  tests  must  be  kept  in  mind. 


TESTS  OF  REINFORCED   CONCRETE 


23 


Num- 
ber of 

Tests 

Type  of  Rod 

Size 
Inchea 

Mixture 

Length 
of  Grip 
Inches 

Maxi- 
mum 
Load 
Pounds 

Bond 
Lb.  Per 
Sq.  In. 

Friction 
Lb.  Per 
Sq.  In. 

6 

Plain  round 

i 

1:3:5| 

6 

3498 

372 

210 

6 

i 

1:2:4 

6 

3893 

412 

227 

6 

1 

1:3:5| 

6 

4170 

355 

227 

4 

1 

1:2:4 

6 

5376 

465 

297 

3 

i 

1:3:5| 

12 

7035 

373 

268 

4 

i 

1:2:4 

12 

7605 

404 

266 

3 

1 

1:3:5| 

12 

9458 

402 

228 

3 

1 

1:2:4 

12 

9736 

414 

223 

3 

Cold  rolled  shafting 

i 

1:3:5! 

6 

2570 

136 

67 

3 

«         a           i( 

i 

1:3:5^ 

6 

1476 

157 

50 

3 

Mild  steel 

ft  x  H 

1:3:5! 

6 

2536 

125 

84 

3 

Round  tool  steel 

3 
4 

1:3:6 

6 

2077 

147 

— 

The  above  table  is  from  Bulletin  No.  8,  University  of  Illinois 
Engineering  Experiment  Station.  In  these  tests  the  rod  was 
imbedded  in  the  whole  thickness  of  the  concrete,  which  was 
such  that  the  elastic  limit  of  the  steel  was  not  exceeded  during 
the  test.  The  concrete  blocks  were  cylinders  6  inches  in  diam- 
eter, and  were,  at  the  time  of  the  tests,  60  days  old. 

The  following  tests  were  made  at  the  University  of  Wisconsin, 
and  were  published  in  Engineering  Record,  Vol.  LVII,  p.  798. 
They  were  made  on  beams  in  which  the  lower  reinforcing  bar 
was  imbedded  only  a  short  distance  from  each  end,  leaving  the 
middle  portion  exposed.  The  stress  in  the  rod  was  computed 
from  the  observed  deformation. 


Mixture 

Age 
Days 

Diameter 
of  Rod 
Inches 

Bond 
Lb.  Per  Sq.  In. 

Bond  by 
Direct  Tension 
Lb.  Per  Sq.  In. 

Ratio 

1:2:4 

60 

1 

278 

394 

1.42 

1:2:4 

60 

I 

286 

455 

1.54 

1:2:4 

60 

1 

276 

— 



1:2:4 

60 

264 

502 

1.90 

1:2:4 

60 

1 

163 

487 

2.99 

1:2:4 

28 

I 

236 

— 



1:2:4 

28 

I 

266 

— 

1.76 

It  will  be  noted  in  the  former  table  that  the  unit  bond  strength 
was  practically  the  same  whether  the  bar  was  imbedded  6  inches 


24  REINFORCED  CONCRETE 

or  12  inches,  that  it  was  greater  with  the  richer  concrete,  and 
was  much  affected  by  the  surface  of  the  steel.  The  latter  table 
shows  clearly  the  marked  difference  in  results  from  the  two 
methods  of  making  the  tests.  Nowhere  is  there  any  marked 
difference  due  to  size  of  the  rod. 

With  the  intention  of  increasing  the  bond,  many  specially 
rolled  bars  of  varying  cross-section  have  been  put  upon  the 
market.  Some  of  these  are  shown  in  the  advertising  pages  of 
all  engineering  periodicals.  New  designs  are  being  made  con- 
stantly and  as  many  are  being  discontinued. 

Tests  made  with  these  bars  show  considerably  higher  bond 
strength  due  to  the  mechanical  bond  produced  by  the  inequali- 
ties of  the  cross-section,  especially  after  the  first  slip  has  taken 
place. 

In  Europe  it  is  a  common  practice  to  bend  the  rods  into  hooks 
at  the  ends  as  shown  in  Fig.  8.  The  corrugated  bars  do  not 
find  the  same  favor  there  as  in  this  country,  the  contention 
being  that  there  is  a  greater  liability  of  splitting  the  concrete 
than  when  plain  reinforcement  is  used.  Such  undoubtedly  is 

the  tendency,  as  the  concrete 
that  fills  the  indentations  is 
sheared;  but  this  action 
takes  place  only  after  plain 
bars  would  have  slipped. 

FIG  g  The  effect  of  the  hook  is  of 

a   somewhat   similar  nature 

in  that  it  prevents  sudden  failure  after  initial  slip  has  taken 
place.  As  will  be  shown,  it  is  sometimes  impracticable  to 
secure  a  length  of  grip  sufficient  to  develop  the  strength  of  the 
bars,  and  in  such  event  the  hooked  end  or  other  form  of  anchor- 
ing is  effective. 

The  above  tables  show  that  bond  strength  in  beams  is  about 
250  pounds  per  square  inch  of  contact.  A  reasonable  factor 
of  safety  would  indicate  that,  for  plain  bars,  the  working  strength 
should  be  taken  at  about  65  to  80  pounds.  If  deformed  bars 
be  used  the  bond  may  be  assumed  to  be  100  pounds  per  square 
inch. 

Let  St  be  the  working  tensile  strength  of  the  steel,  d  the 
diameter  or  thickness  of  the  bar,  and  u  the  unit  bond  strength. 
Then,  to  develop  the  strength  of  the  steel,  whether  round  rods 


TESTS  OF  REINFORCED   CONCRETE  25 

or  square  bars,  the  length  of  imbedment  or  grip  should  be  deter- 
mined from  . 

J  TT  di2  St  =  Trdixu  for  rods 

or  di2  St  =  4  di  x  u  for  bars 

diSt 

whence  x  =  -A —  /1 , 

4w  (1) 

where  u  is  the  average  unit  bond  stress  between  the  concrete 
and  the  steel  over  the  length  x.  With  usual  values  this  indicates 
that  the  grip  should  be  about  50  diameters  of  the  rod. 

The  bond  stress  on  the  bar  in  a  certain  length  depends  upon 
the  change  in  tensile  stress  in  the  bar  within  the  same  distance. 
In  the  distance  d  x  the  change  in  the  bending  moment  is  d  M 
and  the  unit  change  is  dM  +  dx  =  V,  according  to  the  prin- 
ciples of  mechanics  of  homogeneous  beams.  For  equilibrium 
the  bending  moment  equals  the  resisting  moment  which  is 
AStj,  in  which  A  is  the  cross-section  of  the  steel,  and  j  is  its 
lever  arm.  Then,  differentiating,  dM  =  AjdSt,  and  by  sub- 
stitution  AdSi  ^  y 

dx        j 

The  first  member  is  the  change  in  tensile  stress  in  the  bar 
per  unit  of  length.  If  the  concrete  and  steel  be  in  contact, 
the  same  stress  must  take  place  in  the  bond  per  unit  of  length, 
which  is  m  o  u,  where  o  is  the  periphery  of  the  bar  and  m  is  the 
number  of  bars  in  the  section,  if  the  bars  be  alike,  or  in  any 
case,  m  o  is  the  sum  of  the  peripheries  of  all  the  bars  in  the  sec- 
tion. Then  mou  =  V  -f-  j  and 

V 

U     =     .  /r)N 

moj  (2) 

In  many  instances  the  thickness  of  a  beam  will  be  governed 
by  this  determination  of  j,  rather  than  by  the  bending  moment. 

The  results  of  some  tests  of  bond  stresses  in  beams  are  given 
on  page- 31. 

Spacing  of  Bars.  See  Report  of  Joint  Committee,  page  149, 
and  also  page  77. 

Flexural  Tests.  In  making  bending  tests  of  beams,  it  is 
usual  to  place  the  equal  loads  on  points  dividing  the  length 
into  three  equal  parts.  The  general  arrangement  is  shown  in 
Fig.  9.  The  advantage  of  this  arrangement  is  that  the  bending 


26  REINFORCED  CONCRETE 

moment  is  practically  uniform  within  the  space  between  the 
loads.     Measuring  devices  are  attached  so  that  the  deformations 

of  the  fibers  at  the  top  and  at  the 
|  bottom,  and  the  vertical   deflec- 

tion of  the  beam  may  be  deter- 
mined. 

The  objects  in  view  in  making 
flexural  tests  are,  (a)  to  learn  the 
facts  concerning  the  first  signs  of 
FIG   9  failure;   (6)  to  note  the  location 

and  direction  of  cracks  that  finally 

produce  failure;  (c)  to  determine  the  conditions  at  the  time  of 
final  or  complete  failure.  When  such  information  is  at  hand, 
formulas  may  be  developed  by  means  of  which  inferences  may 
be  drawn  as  to  the  probable  strength  of  other  unbroken  beams. 
The  Neutral  Axis.  When  a  simple  beam  of  homogeneous 
material  bends,  the  upper  fibers  are  in  compression  and  the 
lower  ones  are  in  tension,  while  between  them  there  is  a  neutral 
plane  or  axis.  These  facts  give  rise  to  the  well-known  bending 
equation  M  =  SI  +  c 

in  which  M  is  the  moment  of  the  external  loads,  S  is  the  unit 
stress  in  the  extreme  fibers  distant  c  from  the  center  of  gravity 
of  the  section,  and  /  is  the  moment  of  inertia  of  the  right  section. 
When  the  beam  is  of  such  composite  character  as  steel  and  con- 
crete, the  neutral  plane  is  neither  at  the  gravity  axis  of  the 
section,  nor  does  it  remain  fixed  during  the  successive  applica- 
tions of  varying  loads.  It  is,  then,  of  first  impor- 
tance that  the  position  of  the  neutral  axis  be 
determined  as  accurately  as  may  be.  This  is 
done  by  means  of  measurements  of  the  deforma- 
tions at  the  compression  edge,  and  at  plane  of 
the  reinforcing  bars.  If  in  Fig.  10  the  deforma- 
tions be  aa'  and  bb'  the  neutral  axis,  o,  is  located 
by  connecting  a'b'  with  a  straight  line  intersect- 
ing ab  at  o.  The  assumption  that  the  plane 
section  ab  becomes  the  plane  section  a'b'  after 
bending  is  called  "  Navier's  hypothesis,"  or  the 
assumption  of  the  conservation  of  plane  sections.  Careful 
measurements  show  some  deviation  from  a  plane,  but,  in 


TESTS  OF  REINFORCED  CONCRETE 


27 


general,  the  above  assumption  seems  to  be  warranted  by  the 
results  of  observed  deformations  under  working  loads.  If 
observations  be  made  at  cc'  and  dd',  another  determination  of 
the  neutral  axis  is  made  by  joining  c'd'.  This  was  done  in 
making  the  tests  noted  below  from  Bulletin  No  1,  page  29, 
University  of  Illinois  Engineering  Experiment  Station.  The 
points  a  and  b  were  11,  and  c  and  d  were  6  inches  apart. 


Distance  of  Neutral  Axis 

•  •  • 

Distance  of  Neutral  Axis 

Applied  Load 

from  Top.     Inches 

Applied  Load 

from  Top.     Inches 

By  a'b' 

By  c'd' 

By  a'b' 

By  c'd' 

1,000 

10.0 

7.7 

12,000 

6.95 

6.6 

2,000 

8.5 

7.35 

14,000 

7.05 

6.75 

3,000 

7.85 

7.9 

16,000 

7.2 

6.9 

4,000 

7.65 

7.4 

18,000 

7.3 

7.0 

5,000 

7.05 

7.2 

20,000 

7.65 

7.5 

6,000 

7.2 

6.75 

22,000 

7.95 

7.7 

7,000 

7.0 

6.75 

23,000 

8.1 

7.9 

8,000 

7.0 

6.6 

23,600 

8.15 

7.9 

10,000 

6.9 

6.55 

~ 

" 

— 

This  table  shows  a  fair  agreement  in  the  two  determinations 
of  the  neutral  axis,  since  very  minute  errors  of  measurement 
produce  appreciable  differences  in  results.  Also,  lack  of  uni- 
formity in  the  concrete  has  the  same 
effect.  The  noticeable  discrepancy  be- 
tween the  two  determinations  for  the 
1000  pound  load  will  be  referred  to  later. 
The  ultimate  load  for  this  beam  was 
24,600  pounds.  If  the  safe  working  load 
be  taken  as  £  to  J  of  the  ultimate,  or 
from  4000  to  6000,  the  two  determina- 
tions do  not  differ  widely. 

In  the  above  table  the  rapid  change  in 
position  of  the  neutral  axis  under  the 
light  loads  is  noticeable.  This  rise  of  the 
neutral  axis  is  shown  in  Fig.  11.  Under  FlG 

light  loads  the  concrete  on  the  tension 

side  is  intact  and  acts  with  the  steel,  consequently  the  deforma- 
tion of  the  steel  is  small,  a  large  proportion  of  the  section  is 


28 


REINFORCED  CONCRETE 


in  compression,  and  hence  the  neutral  axis  is  low,  although  not 
stationary.  If  the  coefficient  of  elasticity  be  2,000,000  =  E, 
and  the  ultimate  strength  of  concrete  in  tension  be  St  =  200 
pounds  per  square  inch,  the  ultimate  unit  elongation  of  the 
concrete  is  .0001.  When  such  loads  have  been  applied  as  will 
cause  a  greater  elongation,  the  concrete  in  tension  becomes 
broken  by  fine,  usually  unnoticeable  cracks,  and  the  steel  takes 
a  greater  part  of  the  tensile  stress.  The  elongation  &i&2  increases 
faster  than  a^a^  and  the  neutral  axis  rises  rapidly.  Additional 
loads  cause  more  and  deeper  cracks  in  the  concrete  and  a  con- 
tinued upward  but  slower  movement  of  the  axis.  These  two 
stages  represent  what  takes  place '  in  a  beam  when  carrying 
rather  less  than  the  safe  working  loads. 

The  action  of  a  beam  under  ordinary  working  loads  repre- 
sents another  stage  during  which  the  position  of  the  neutral 
axis  moves  but  little  and  the  steel  takes  all  the  tension.  The 
following  table,  taken  from  the  same  source  as  that  above, 
shows  the  position  of  the  neutral  axis  during  this  stage. 


Beam   No. 

Per  Cent,  of 
Reinforcement 

Proportionate 
Depth 

Beam  No. 

Per  Cent,  of 
Reinforcement 

Proportionate 
Depth 

21 

0.41 

.34 

14 

1.11 

.46 

19 

0.41 

.36 

5 

0.83 

.42 

16 

0.52 

.375 

28 

1.52 

.53 

17 

0.52 

.37 

13 

0.97 

.45 

27 

1.56 

.53 

20 

0.69 

.44 

9 

0.52 

.34 

2 

0.69 

.39 

15 

0.83 

.41 

7 

0.42 

.33 

10 

0.83 

.43 

3 

0.42 

.31 

22 

1.67 

.57 

29 

1.48 

.52 

4 

1.39 

.47 

— 

— 

— 

Beam  No.  22  in  this  table  is  the  one  from  which  the  previous 
table  was  taken.  It  will  be  noted  that  the  ratio  of  the  area  of 
the  cross-section  of  steel  to  that  of  concrete  was  large:  the  con- 
sequence was  that  the  final  loads  caused  excessive  stresses  in 
the  concrete  while  the  steel  was  not  stressed  above  its  elastic 
limit.  This  represents  the  final,  or  failing  stage,  and  the  neutral 
axis  is  seen  to  fall.  This  final  drop  of  the  axis  does  not  occur 
when  the  reinforcement  is  such  that  failure  takes  place  first  in 


TESTS  OF  REINFORCED  CONCRETE  29 

the  steel.  If  the  ultimate  strength  of  the  concrete  in  com- 
pression be  Sc  =  3000  pounds  per  square  inch,  this  fourth  stage 
will  be  entered  upon  when  the  L  j 

unit  deformation  becomes  .0015 
on  the  compression  side. 


In    Fig.    12    are    shown    the  i 

three  typical  cases  of  beam  fail- 
ure. If  the  component  parts  of 
the  beam  were  of  the  strengths 
assumed,  the  design  might  be 
such  that  the  full  strength  in 


\   V    \ 


compression,  tension,  and  shear  c 

would  be  reached  at  the  same 

time.     Such  a  condition  is  seldom  realized  in  practice,  but  the 

three  causes  of  failure  may  follow  each  other  so  closely  that 

rupture  may  be  attributed  to  the  wrong  one. 

Compression.  In  a  the  proportion  of  steel  is  sufficient  to 
develop  the  full  strength  of  the  concrete  before  itself  is  stressed 
beyond  the  elastic  limit.  Failure  in  this  way  takes  place  slowly, 
and,  in  actual  use  in  buildings,  ample  warning  is  given  that 
supports  must  be  provided  and  repairs  made.  An  area  of  steel 
cross-section  from  1  per  cent,  to  1.5  per  cent,  of  that  of  the 
beam  will  usually  be  sufficient  to  prevent  failure  by  tension. 

Tension.  The  cost  of  steel  is  often  such  in  comparison  with 
that  of  concrete  that  it  is  not  economical  to  use  enough  of  the 
former  to  develop  the  ultimate  strength  of  the  concrete  in  com- 
pression. Test  beams  then  may  show  open  cracks,  as  in  6, 
Fig.  12,  on  the  under  side;  they  begin  where  the  bending  moment 
is  large,  the  exact  location  being  determined  by  the  presence  of 
a  section  weaker  than  others  having  the  same  or  greater  bending 
moments.  Failure  by  tension  occurs  rapidly  when  the  elastic 
limit  of  the  steel  is  exceeded. 

Diagonal  Tension.  The  crack  shown  in  c,  Fig.  12,  usually 
begins,  in  test  beams  with  two  loads,  between  one  load  and  the 
nearer  support,  and  continues  upward  and  toward  the  middle 
of  the  span.  This  is  a  diagonal  tension-failure,  sometimes  called 
shear,  and  is  a  combination  of  shear  and  tension  which  may 
cause  rupture  before  the  steel  is  stressed  beyond  its  elastic  limit 
or  the  concrete  in  compression  is  broken.  Failure  from  this 
cause  occurs  in  beams  having  a  depth  large  in  comparison  with 


30  REINFORCED  CONCRETE 

the  length,  since  in  such  cases  a  high  vertical  shear  is  developed. 
The  unit  diagonal  tension  in  homogeneous  beams  is  given  by 
the  formula  t  =  %S  +  \/\  S2  +  v2,  in  which  S  and  v  are  the 
unit  stresses  due  to  horizontal  tension  and  vertical  shear 
respectively.  The  direction  of  the  diagonal  tension  is  expressed 
in  tan  2  0  =  2  v  -r-  S  where  0  is  the  angle  between  the  direction 
of  the  stress  and  the  horizontal.  At  the  neutral  axis  S  is  zero, 
and  9  reduces  to  45°,  while  t  becomes  equal  to  v. 

Diagonal     tension     is 

provided  for  by  bending 

up  the  ends  of  the  lon- 

~ ; —  —     gitudinal   bars   that  are 

not  needed   far  beyond 

the  middle  of  the  beam,  and  by  the  insertion  of  vertical  rods 
or  U-shaped  forms  called  stirrups.  The  bent  up  rods  are  shown 
at  a  and  the  stirrups  at  6,  in  Fig.  13. 

Bond  and  Shear.  For  ordinary  loads  and  for  beam  sizes 
necessary  to  provide  against  other  varieties  of  failure,  shear  is 
seldom  a  controlling  factor  in  the  design.  If  the  area  of  cross- 
section  of  the  steel  be  about  f  of  1  per  cent,  of  that  of  the  beam, 
the  usual  beam  formula  applies  to  reinforced  concrete  beams. 
As  will  be  shown  later,  for  homogeneous  rectangular  beams  the 
maximum  unit  shear  is  f  the  average.  Then,  for  uniform  load, 
if  C  be  the  unit  stress  in  the  concrete  and  the  neutral  axis  be 
in  the  middle, 

C  =  6  M  -f-  bd2  =  f  wl2  +  bd2 
and  v  =  %V  +  U  =  lwl  +  bd 

hence  v:  C  =  d:  I. 

According  to  Table,  page  18,  v  is  more  than  half  the  com- 
pressive  strength  of  concrete,  so  it  is  unlikely  that  a  beam  will 
break  by  direct  shear.  Of  course  the  above  assumptions  as  to 
the  amount  of  reinforcement  and  load  are  not  general,  but 
under  usual  conditions  the  conclusion  is  correct.  On  page  79 
it  will  be  shown  that  l/d  may  be  even  less  than  C/v  before  the 
shearing  strength  is  exceeded  by  the  shearing  stress. 

The  following  table  of  bond  and  shear  stresses  in  a  beam  is 
from  Bulletin  No.  4  University  of  Illinois  Engineering  Experi- 
ment Station.  In  the  expressions  for  v  and  u,  j  is  the  vertical 
distance  from  the  center  of  gravity  of  the  compressive  stresses 


TESTS  OF  REINFORCED   CONCRETE 


31 


to  that  of  the  tensile  stresses.  In  a  homogeneous  beam,  j  =  f  d, 
making  v  =  I  V  -?-  bd  as  used  above.  In  a  reinforced  concrete 
beam,  j  is  variable,  depending  upon  the  position  of  the  neutral 
axis.  The  area  of  contact  between  a  reinforcing  bar  and  the 
concrete  is  o,  and  the  number  of  bars  is  m. 


Reinforcement  % 

Vertical  shear 
v  =  V-*-  b  j 
Lb.  per  Sq.  In. 

Bond 
u  =  VH-  m  o  j 
Lb.  per  Sq.  In. 

2.21  mild  steel 

130 

110 

2.21     " 

124 

106 

2.21     ' 

t 

137 

117 

1.66     ' 

I 

120 

135 

1.66     ' 

i 

104 

116 

1.60     ' 

i 

117 

143 

1.84     '       ' 

109 

112 

1.10  tool  st< 

el 

95 

161 

1.10 

72 

123 

1.10 

66 

112 

1.10 

107 

181 

1.10 

73 

124 

1.10 

73 

124 

1.66 

101 

114 

1.66 

126 

143 

1.66 

« 

107 

120 

The  concrete  was  of  1:3:6  mixture.  It  will  be  noted  the 
values  of  v  are  much  smaller  than  the  ultimate  strength  as 
given  on  page  18. 

Repetition  of  Stress.  It  is  well  known  that  an  enormous 
number  of  repetitions  of  a  comparatively  small  load  will  cause 
fracture  in  a  piece  of 
metal  capable  of  sustain- 
ing a  static  load  many 
times  as  great.  An  ac- 
count of  experiments  to 
determine  the  effect  of 
repetition  of  loads  on 
concrete  is  given  in 
Trans.  Am.  Soc.  C.  E., 
Vol.  LVIII.  The  sum- 
mary is  shown  in  Fig.  14. 
It  is  seen  that  when  the  load  applied  is,  for  example,  60  per  cent. 


S  a 

li'80 

1 


.40 


s 

^ 



8000          16000         24000         &.JOO 
Number  of  Repetitions 
Producing  Failure 

FIG.  14. 


32 


REINFORCED  CONCRETE 


of  the  ultimate,  4000  applications  cause  failure;  when  it  is  55 
per  cent,  about  8000  applications  are  required.  If  loads  repre- 
senting usual  factors  of  safety,  as  4  or  5,  be  considered,  the  num- 
ber of  repetitions  required  to  cause  failure  is  comparable  with 
the  number  necessary  in  the  case  of  iron. 

In  Engineering  Record,  Vol.  58,  page  90,  is  an  account  of 
repeated  loadings  made  at  University  of  Pennsylvania  on  con- 
crete beams.  Under  loads  causing  25  per  cent,  of  the  ultimate 
stresses  the  set  remained  practically  constant  for  360,000  applica- 
tions. The  amount  of  set  increased  materially  under  heavier 
loads,  and  500,000  repetitions  of  a  40  per  cent,  lead  did  not 
seem  to  affect  the  ultimate  strength  of  the  beam. 

Fig.  15  shows  the  stress-strain  diagram  of  a  beam  as 
reported  in  Bulletin  No.  14,  Univ.  of  111.  Engineering  Experi- 
ment Station. 
The  mixture 
was  1:3:5.5 
concrete;  the 
span  was  12 
feet,  the  width 
was  8  inches, 
and  the  effec- 
tive depth  was 
10  inches.  It 
will  be  noted 
that  the  elastic 


Deformation  per  Unit  of  Length 
.0004  0.001  .0016 


Span  12  ft.  Reinforcement  0.98  % 
Concrete  1-3-5H 


FIG.  15. 
deformation  remained  quite  constant  throughout. 

Expansion  and  Contraction.  In  structures  containing  ex- 
tended areas  of  concrete,  allowance  must  be  made  for  expansion 
and  contraction,  which  occur  from  two  causes.  Like  most 
building  material,  concrete  expands  as  the  temperature  rises, 
and  contracts  as  it  falls.  Experiments  made  to  determine  the 
rate  of  change  in  length  of  concrete  due  to  change  of  temperature 
have  not  been  as  numerous  or  as  conclusive  as  could  be  desired. 
However,  the  tests  seem  to  indicate  that  the  coefficient  of  ex- 
pansion is  between  .0000055  and  .0000065,  and  that  the  mean 
of  these  may  be  accepted  as  the  average  value.  The  coefficient 
of  expansion  of  steel  is  .0000065  to  .0000067,  which  is  so  nearly 
like  that  of  concrete  that  only  small  stresses  result  from  the 
difference. 


TESTS  OF  REINFORCED  CONCRETE  33 

Concrete,  in  hardening  under  water,  expands  somewhat, 
usually  from  .0002  to  .0005  of  its  length,  while  it  contracts  from 
.0003  to  .0005  of  its  length  if  kept  in  air.  For  the  same  mixture 
the  change  in  air  is  generally  more  marked,  and  as  between 
different  mixtures,  the  richer  suffers  the  greater  changes.  These 
changes  may  continue  slowly  for  months,  but  about  half  the 
whole  will  be  effected  in  perhaps  a  week.  As  most  reinforced 
concrete  structures  are  built  within  forms  that  retain,  for  one 
or  two  weeks,  some  of  the  extra  water  used  in  mixing,  the  expan- 
sion or  contraction  should  often  not  be  more  than  the  smaller 
values  given  above. 

Cracks  are  sure  to  appear  unless  they  are  uniformly  distrib- 
uted by  means  of  reinforcing  metal,  or  contraction  joints  are 
provided  wherein  the  contraction  may  take  place  without  being 
noticeable.  The  unsightly  appearance  of  a  wall  cracked  by 
contraction  in  the  concrete  should  be  a  sufficient  reason  for 
the  prevention  of  the  same,  but  the  stability  of  the  structure 
may  be  endangered  if  the  openings  allow  entrance  of  water  or 
moisture  in  sufficient  quantities  to  cause  the  steel  to  rust. 

Weight  of  Reinforced  Concrete.  Experiments  to  determine 
the  weight  of  concrete,  made  at  the  Watertown  Arsenal,  are 
published  in  "Tests  of  Metals,"  1897-99  and  1903-4.  The  con- 
clusions are  that  dense,  well  made  1:2:4  concrete,  when  dry, 
weighs  110,  145,  150,  152,  and  155  pounds  per  cubic  foot,  accord- 
ing as  the  aggregate  is  cinder,  sandstone,  limestone,  gravel,  or 
trap  rock.  A  1:3:6  concrete  of  trap  weighs  150,  of  gravel,  145, 
and  of  cinder,  105  pounds  per  cubic  foot.  One  per  cent,  of  steel 
adds  about  3.5  pounds  to  these  weights. 

The  assumed  weight' of  reinforced  concrete  is  usually  150 
pounds  per  cubic  foot. 

The  Reinforcing  Metal.  Most  of  the  steel  used  as  reinforce- 
ment is  in  the  form  of  bars  of  round  or  square  cross-section. 
Flat  shapes  do  not  effect  as  reliable  bond,  otherwise  they  would 
have  some  advantage  over  the  square  section  in  that  the  centers 
may  be  placed  farther  from  the  neutral  axis.  Various  bars  of 
deformed  cross-section,  designed  to  secure  mechanical  bond,  are 
widely  advertised.  Many  of  these  shapes  are  popular  and 
effective,  and  are  sold  at  prices  not  much  above  plain  bars. 
Woven  wire  and  thin  plates,  punched  so  as  to  secure  a  bonding 
section,  are  used  in  thin  slabs  with  good  effect.  The  sizes  of 


34 


REINFORCED  CONCRETE 


bars  vary  by  iV-inch  increments,  from  J  inch  to  an  inch,  and  then 
by  f-inch  increments,  to  IJ-inch  diameters.  Occasionally  2-inch 
bars  are  used,  but  they  are  bent  with  some  difficulty. 

Qualities  of  Steel.  Specifications  for  reinforcing  steel  are 
given  on  page  155.  In  general  such  steel  is  of  two  varieties, 
"  medium  steel  "  and  "  high  elastic  limit  steel."  A  comparison 
between  the  two  is  shown  below. 


Ultimate  Tensile 
Strength 

Elastic  Limit 

Elastic  Deforma- 
tion 

Pounds  per  Square  Inch 

Per  Unit  of  Length 

Medium  steel 
High  elastic  limit  steel 

60,000  to    70,000 
80,000  to  100,000 

35,000  to  40,000 
50,000  to  60,000 

0.0010  to  0.0013 
0.0015  to  0.0020 

The  coefficient  of  elasticity  of  steel  is  assumed  to  be  30,000,000  pound 
inches.  The  coefficient  of  expansion  of  steel  is  usually  taken  at  .0000065  for 
each  degree  Fahr.  of  change  of  temperature. 

For  other  specifications  for  steel  see  report  of  Committee  A-l  to  the 
American  Society  of  Testing  Materials,  June  28,  1911.  This  report  was 
approved  by  the  meeting  for  submission  to  letter  ballot. 

Proportioning  Concrete.  The  exact  amounts  of  cement,  sand, 
and  broken  stone  necessary  to  make  a  cubic  yard  of  concrete 
have  not  been  definitely  determined.  The  reason  is  that  so 
many  factors,  such  as  sharpness  of  sand  and  of  pieces  of  the 
aggregate,  percentage  of  voids  in  the  same,  dampness  of  the 
sand,  manner  of  packing,  and  measuring  the  cement  and  amount 
of  water  used,  make  a  concise  statement  impossible.  A  valu- 
able article,  with  discussions,  on  the  effect  of  varying  the  pro- 
portions in  concrete,  is  published  in  Trans.  Am.  Soc.  C.  E., 
Vol.  LIX,  December,  1907. 

Quantities  Required  for  a  Cubic  Yard  of  Concrete.  The 
table  below  is  from  various  sources,  and  is  supposed  to  represent 
common  practice  when  cement  is  measured  by  bags  of  96  pounds, 
and  loose  sand  and  run  of  crusher  broken  stone  are  used. 


TESTS  OF  REINFORCED  CONCRETE 


35 


Proportion 
by  Parta 

Volumes  in  a  Cubic  Yard  of  Concrete 

Volumes  of  Sand  and  Stone  to  a  Unit 
of  Cement 

Cement, 
Bbl. 

Sand, 
Cu.   Yd. 

Stone, 
Cu.   Yd. 

Cement, 
Bbl. 

Sand, 
Cu.  Ft. 

Stone, 
Cu.  Ft. 

:2:3 

1.65 

0.45 

0.70 

7.6 

11.4 

:2:4 

1.40 

0.40 

0.80 

7.6 

15.2 

:2:5 

1.20 

0.35 

0.85 

7.6 

19.0 

:3:5 

1.10 

0.50 

0.75 

11.4 

19.0 

:3:6 

1.00 

0.45 

0.85 

11.4 

22.8 

:3:S 

0.80 

0.35 

0.90 

11.4 

30.4 

1:4:8 

0.75 

0.40 

0.85 

1 

15.2 

30.4 

The  sand  and  broken  stone  are  often  measured  by  the  wheel- 
barrow load,  in  which  case  a  box  containing  a  cubic  foot  should 
be  frequently  used  to  verify  the  loads. 

Fuller's  Rule.  A  simple  rule  is  proposed  by  Mr.  W.  B.  Fuller, 
which  gives  fairly  accurate  proportioning  for  packed  cement, 
sand,  and  broken  stone  that  has  about  40  per  cent  of  voids. 
Let  c,  s,  and  g  be  number  of  parts  of  cement,  sand,  and  gravel 
or  broken  stone  respectively.  Let  C,  S,  and  G  be  the  number 
of  barrels  of  packed  cement,  the  number  of  cubic  yards  of  loose 
sand  and  gravel  or  broken  stone  respectively,  required  for  a 
cubic  yard  of  concrete,  then 

11 


C  = 


+  g 


~ 


If  stone  be  screened  to  uniform  size  add  5  per  cent.,  and,  for 
well-graded  stone,  deduct  5  per  cent,  from  all  quantities. 

Depth  of  Concrete  below  Steel.  There  are  two  reasons  for 
covering  the  rods  in  the  tension  side  of  a  beam  with  concrete: 
without  this  covering  the  bond  would  be  imperfect;  and  by  it 
protection  is  given  from  fire  and  from  moisture.  The  recom- 
mendations of  the  Joint  Committee  (p.  144)  are  that  the  thick- 
ness of  the  concrete  outside  the  rods  be:  for  slabs,  1  inch;  for 
beams,  1|  inches,  and  for  girders,  2  inches. 


36  REINFORCED  CONCRETE 

/ 

PROBLEMS 

2.  Compute  the  number  of  cubic  yards  in  a  concrete  pier  4  feet  by 
16  feet  on  top  and  22  feet  high.  The  four  sides  have  a  batter  of  an  inch 
to  the  foot. 

3.  Compute  the  amount  of  cement,  sand,  and  broken  stone  required 
to  build  the  pier  in  Problem  2,  the  mixtures  being  1:  2:  5,  1:  3:  6,  and 
1:4:8. 

4.  With  cement  at  $1.65  a  barrel,  sand,  $0.75,  and  broken  stone, 
$1.50  per  cubic  yard,  which  of  the  mixtures  in  Problem  3  is  the  cheaper 
if  the  strength  according  to  Table,  page  16,  be  considered? 


CHAPTER   IV 


ANALYSIS  OF  STRESSES 

Stresses  in  Beams.  Any  beam,  whether  homogeneous  or 
otherwise;  will,  when  loaded,  be  stressed  in  tension,  compression, 
and  shear.  The  horizontal  tension  and  compression  stresses 
unite  with  the  vertical  shear  to  produce  resultant  diagonal 
tension  and  diagonal  compression  stresses.  In  beams  of  material 
that  has  about  equal  tensile  and  compressive  strength,  the  diag- 
onal stresses  are  not  so  vital  as  in  concrete  beams.  These  stresses 
will  be  considered  in  order. 

Beam  Theories.  Navier's  hypothesis  pertains  to  beams  of 
all  materials,  and,  as  explained  in  page  26,  holds  good  for  work- 
ing loads  on  those  of  concrete.  In  homogene- 
ous beams,  where  the  tensile  and  compressive 
elasticities  are  alike  and  Hooke's  law  applies, 
the  stress  diagram  is  like  Fig.  16,  where  a'b'  is 
a  straight  line,  and  o  is  in  the  middle  of  ab. 
When  concrete  is  stressed  above  safe  limits,  as 
^  6'  the  coefficient  of  elasticity  of 

concrete  is  not  constant  for  all 
stresses,  the  stress  diagram  is 
not  bounded  by  straight  lines  a< 
as  before,  but  is  represented  by 
Fig.  17.  For  equilibrium  the  areas  066'  and  oaa' 
must  be  equal,  so,  if  aa'  is  smaller  than  bb',  the 
neutral  axis  is  above  the  middle.  The  stress  is 
not  proportional  to  the  deformation,  and  if  com- 
puted by  the  ordinary  beam  formula,  or  from 
measured  deformations  with  the  assumption  that 
a'b'  is  straight,  high  values  of  modulus  of  rup- 
ture will  be  obtained  as  was  shown  on  page  27. 
When  steel  is  introduced  to  take  tension  in  the  beam,  it 
is  capable  of  much  greater  elongation  without  breaking  than  is 
concrete.  For  this  reason  the  concrete  on  the  tension  side  is 

37 


FIG.  16. 


FIG.  17. 


38 


REINFORCED  CONCRETE 


broken  at  frequent  intervals,  the  cracks  extending  inward  toward 
the  neutral  axis.  While  they  do  not  reach  quite  to  that  plane, 
the  tensile  strength  of  the  concrete  is  so  nearly  destroyed  before 
the  safe  working  strength  of  the  steel  is  developed,  that  it  is 
usual  to  assume  the  steel  taking  all  the  tension.  The  cracks 
are  most  numerous  and  deepest  where  the  bending  moments 
are  greatest  or  where  the  concrete  happens  to  be  weakest.  The 
concrete  may  be  intact  and  able  to  resist  shear  and  diagonal 
tension  near  the  ends  of  the  beam  where  the  shear  is  greatest. 

Fig.  18  shows  the  stress  diagram  on  the  left  where  the  line 
cc'  represents  the  ultimate  compressive  strength,  and  W  the 
stress  under  any  given  loading.  If  c  and  s,  in  Fig.  18  6,  be 

measured,  the  location  of  the 
neutral  axis  is  known,  and  the 
total  tensile  stress  in  the  steel 
is  computed  from  E  =  S  -f-  s, 
in  which  E  is  the  coefficient  of 
elasticity,  S  is  the  unit  stress, 
and  s  the  unit  elongation  of  the 
steel.  Then,  for  equilibrium, 
the  total  compression  and  ten- 
sion are  equal,  and  the  area  of 
obb'  is  determined,  hence  the 
nature  of  the  curve  cib'c'  will  be  known  if  a  sufficient  number 
of  points,  as  &',  be  fixed. 

For  working  stresses,  as  W,  Fig.  18,  the  line  ob'  may  be  taken 
as  straight  without  material  error  in  results. 

For  loads  that  cause  the  destruction  of  a  beam,  the  straight 
line  assumption  certainly  does  not  hold  true.  Repeated  observa- 
tions of  the  stress  diagrams  indicate  that  ob'c'  is  very  nearly  a 
parabola  whose  origin  is  at  cf  when  cc'  is  the  ultimate  strength 
of  concrete.  There  may  be  some  curve  that  fits  the  plotted 
points  even  better  than  does  the  parabola;  but  no  other  as 
simple  and  well  known  curve  seems  available.  Hence,  the 
parabolic  stress  diagram  for  ultimate  loads  is  universally  adopted 
in  analyzing  the  stresses  in  beams  tested  to  destruction. 

The  Straight  Line  Analysis.  In  the  following  pages  the 
nomenclature  is  as  given  below,  unless  otherwise  stated. 

/  =  length  of  beam. 
b  =  breadth  of  beam. 


FIG.  18. 


ANALYSIS  OF  STRESSES 


39 


d  =  distance  from  center  of  tensile  steel  area  to  compression 
side  of  beam,  or  the  effective  depth. 

h  =  entire  depth  of  beam. 

A  =  area  of  tensile  steel  in  section  considered. 

A'  =  area  of  compressive  steel  in  section  considered. 

p  =  A  -5-  bd  =  proportion  of  steel,  in  section. 

p  =  A  -r-  bh,  when  the  whole  section  is  in  compression. 

k  =  distance  of  neutral  axis  from  compressive  side. 
St  =  tensile  unit  stress  in  the  steel. 
Sc  =  compressive  stress  in  the  steel. 

C  =  compressive  stress  in  the  concrete. 

r  =  St  -s-  C. 

r'  =  Se  +  C. 

Es  =  coefficient  of  elasticity  of  steel. 
Ec  =  coefficient  of  elasticity  of  concrete. 

s  =  unit  elongation  of  steel. 

c  =  unit  elongation  of  concrete. 

n  =  Es  -=-  Ec. 

M  =  bending  moment  due  to  loads. 
Ni  Nt,  etc.,  =  constants  in  stress  formulas. 
V  =  vertical  shear. 

v  =  unit  shearing  stress. 

/  =  moment  of  inertia  with  respect  to  gravity  axis. 

j  =  vertical  distance  between  points  of  application  of  hori- 
zontal tensile  and  compressive  stresses. 
b'  =  breadth  of  web  of  T-beam. 

t  =  depth  of  slab  of  T-beam. 
Ib.  sq.  in.  =  pounds  per  square  inch. 

Ib.  in.  =  product  of  pounds  by  inches. 

Fig.  19  represents  the  stresses  in  a  beam  due  to  bending.     For 
equilibrium  the  sum  of  all  horizontal  forces  is 

zero,  so 

ASt  =  \  bCk  (3) 

Since  the  sum  of  all  the  moments  is  zero 

ASt  (d  -  k)  +  J  bCk2  =  M  (4) 

Since  a  plain  section  remains  plain  after  bending 
d-k       St        C       StEc 


k 


or 


k  = 


Ec      CES 

dn 


(5) 


From  (6)  and  (4)  A  = 


r  -\-  n 
bdn 


3  (n  +  r)M 

2  r  (r  +  n)      dSt  (3  r  +  2  n) 


(6) 

(7) 


40  REINFORCED  CONCRETE 

'-270T+7T  •  (8) 

From  6  and  8,  k  =  2  prd  (9) 

Putting  the  value  of  r  in  (8)  into  (9) 


=  d  (-  up  ±  V2  np+  r*2p2)  (10) 


The  positive  sign  of  the  radical  is  to  be  used. 
From  (7) 


since  in  Fig.  19,  3  =  d  —  \  k. 

An  expression  for  the  depth  is  obtained  by  solving  (7)  for  d 
3  (n  +  r)  M 


A&    (3r  +  2rc) 
Substituting  the  first  value  of  A  in  (7) 


(13) 


_     6  r  (r  +  n)«  M  ,     , 

" 


Since  S  =  rC 
From  which 


_      6  (r  +  n)»  M  ,     , 

~ 


3  r  +  2  n 


(16) 


When  Ni  is  |  this  becomes  the  ordinary  beam  formula, 
M  =  SI  -:-  c,  where  c  is  the  distance  from  the  gravity  axis  to 
the  extreme  fiber. 

In  Plate  1  are  plotted  values  of  Ni  for  three  values  of  n. 
Crossing  these  curves  are  lines  representing  values  of  p  taken 
from  equation  (8),  while  values  of  r  are  given  by  abscissas. 

It  will  be  noted  that  of  the  factors  involved,  N,  r,  p,  and  n, 
only  two  may  be  arbitrarily  assumed  at  once,  as  the  others  are 
dependent  upon  these.  For  example,  if  S  be  15,000  and  C  be 


ANALYSIS  OF  STRESSES 


41 


Diagram  for  Rectangular  Beams 


PLATE  I. 


42  REINFORCED  CONCRETE 

500  pounds  per  square  inch,  r  is  30;  then  if  n  be  15,  p  and  N 
are  0.0056  and  0.148,  and  cannot  be  otherwise  assumed. 

Again,  if  the  loads  p  and  beam  dimensions  be  known,  for 
usual  limits  of  n  (10  to  20),  NI  must  be  between  0.123  and 
0.157,  while  r  will  lie  between  27  and  36;  and  if  these  values, 
with  a  safe  working  stress  for  C,  give  M  less  than  that  computed 
from  the  loads,  the  beam  is  overloaded  or  too  small.  For 
example,  a  beam  is  12.5  feet  long,  b  =  12  inches,  d  =  18  inches, 
and  p  is  .008.  If  the  concrete  be  of  good  quality  the  value  of 
n  may  be  taken  as  15.  It  is  required  to  find  the  safe  load  per 
linear  foot  for  this  beam.  In  the  diagram,  for  n  —  15  and 
p  =  .008,  NI  is  .167,  and  r  is  24.  If  the  concrete  be  capable  of 
sustaining  a  stress  of  600  pounds  per  square  inch,  the  steel 
should  carry  600  X  24  =  14,400,  as  it  usually  can.  With  these 
assumptions 

M  =  .167  X  600  X  12  X  182 

=  398570  Ib.  in. 
12  X  1/8  wl2  =  21640 

w  =  1730  pounds  per  linear  foot. 

Excessive  Reinforcement.  As  a  general  proposition  it  is  not 
economical  to  use  a  greater  amount  of  steel  than  is  indicated 
by  the  intersection  of  the  ordinate  through  r  and  the  curve  of 
n  in  Plate  I.  For  example,  let  the  safe  strength  in  pounds  per 
square  inch  of  steel  and  concrete  be  respectively  16,000  and 
500,  and  let  the  percentage  of  steel  be  1.0.  According  to  the 
diagram,  considering  p  and  n,  r  is  21,  and  hence,  if  C  be  500, 
St  is  only  10,500.  The  difference  between  16,000  and  10,500 
is  paid  for,  but  is  not  used  unless  500  for  C  be  exceeded.  The 
value  of  Ni  is  seen  to  be  .180,  and  the  bending  moment  is 

M  =  .180  X  500  X  bd2 

For  the  assumed  values  of  C  and  St,  r  =  32  and  NI  is  .144 
or  M  =  .144  X  500  X  b'd'z 

To  make  these  values  of  M  identical,  d'2  may  be  changed  in 
the  ratio  of  .180  to  .144,  and  the  new  area  of  cross-section  will 
be  1.12  of  the  first,  while  but  half  as  much  per  cent,  of  steel  is 


ANALYSIS  OF  STRESSES  43 

needed.     The  cost  of  the  two  designs  would  then  be  indicated  by 

1.00  bdx  +  .01  bdy 
and  1.12  bdx  +  .QQ56bdy 

when  x  is  the  cost  of  concrete  and  y  that  of  the  steel.  If  the 
steel  be  worth  50  times  as  much  as  concrete  per  unit  of  volume, 
the  beam  with  1  per  cent,  of  reinforcement  costs  1.07  as  much 
as  that  with  |  per  cent,  reinforcement,  greater  depth  and  the 
same  strength. 

The  questions  of  economical  design  will  be  taken  up  again 
in  Chapter  V,  but  this  example  serves  to  illustrate  the  fact 
that  the  strength  of  a  beam  depends  largely  upon  its  depth. 

Under  Reinforcement.  Assuming  the  same  values  of  C  and 
S  as  above,  makes  r  =  32  and  p  =  0.5  per  cent.  If  the  beam 
be  constructed  with  but  .25  per  cent,  of  steel,  r  will  be  about 
48.  If  St  be  16,000,  C  must  be  16,000  -*•  48  =  333  pounds  per 
square  inch.  On  the  other  hand,  if  500  pounds  unit  stress  be 
developed  in  the  concrete,  the  steel  will  be  carrying  a  unit  load 
of  24,000  pounds.  Hence,  if  it  be  desired  to  adhere  strictly 
to  the  assumed  safe  unit  stresses,  p  must  be  exactly  as  given 
in  the  Plate  I. 

PROBLEMS 

5.  A  simple  beam  12  inches  by  16  inches  by  18  feet  carries  two  loads 
of  2000  pounds  each,  at  4  feet  and  9  feet  from  the  end,  in  addition  to  its 
own  load.     What  percentage  of   steel  is  required  that  the  unit  stress 
in  the  concrete  be  not  over  550  pounds  per  square  inch?    What  will 
be  the  unit  stress  in  the  steel? 

6.  A  simple  beam  of  1:3:  6  concrete  has  a  span  of  16  feet.     If  the 
width  be  10  inches  and  p  be  6  per  cent.,  what  should  be  the  depth  to  sus- 
tain a  uniform  load  of  800  pounds  per  linear  foot  ?    What  should  h  be? 
What  change  in  depth  and  of  p  would  be  allowable  if  a  1:2:4  concrete 
be  used? 

7.  If  the  reinforcement  in  Problems  5  and  6  be  of  three-quarter  inch 
square  bars,  how  far  from  the  middle  of  the  beams  is  it  necessary  that 
all  these  bars  extend  to  provide  for  the  bending  moment? 

8.  If  2.1  square  inches  of  steel  be  required  in  a  breadth  of  12  inches, 
what  size  of  round  bars  may  be  used  to  most  nearly  conform  to  the  rules 
given  on  page  150,  and  to  the  assumed  area? 

Extending  Plate  I.  It  may  happen  that  a  larger  scale  is 
desired  for  this  plate,  or  that  it  is  to  be  used  for  a  concrete  of 


44 


REINFORCED  CONCRETE 


cinders  for  which  n  may  be  as  high  as  30.  In  (8),  if  p  and  n 
be  asumed  a  value  of  r  can  be  found.  With  this  value  of  r 
N  is  found  from  (15).  For  example,  let  it  be  required  to  find 
the  intersection  of  p  =  1  per  cent,  and  n  =  30.  From  (8)  r 
is  found  to  be  26.5,  and  from  (15)  N  is  0.219.  In  like  manner 
points  on  the  lines  indicating  the  values  of  p  are  plotted  and 
all  are  connected  by  a  smooth  curve. 

Double  Reinforcement.  Instead  of 
increasing  the  depth  to  procure  more 
strength,  it  is  sometimes  expedient  to  add 
steel  on  the  compressive  side.  In  Fig.  20 
compressive  steel  of  A'  is  d'  from  the  top. 
For  equilibrium 


M  = 


and  giving  to  ASt  this  value,  or  taking 
center  of  moments  at  the  tensile  steel, 


(18) 


as  a  vertical  section  remains  plane  during  bending 

dn 


k  = 


and 


r  +  n       n  —  r' 

&   f          f\ 

T,  (n  -  r')  -  n 


(19) 
(20) 


Substituting  these  values  of  k  and  r  in  (18), 


M  =  Cb#l"r,  +  *"mn  +  p'r' 


-  I     d  -d 


6  (r  + 


n 


=o 

d    ) 


=  Cbd^lN, 
=  NzCbd2 


I    I 

pr 


n    — 


d 


(21) 
(22) 


It  will  be  noted  that  (22)  and  (16)  are  identical  when  p' 
is  zero. 

In  (18)  A  and  A'  may  be  stated  in  terms  of  p  and  p/  whence 

Jc_       p'r'       p'(k-  d'} 
P  ~  2  dr  "*     r  rk 


ANALYSIS  OF  STRESSES  45 

and,  substituting  the  value  of  k,  above 


From  the  two  values  of  k  the  relation  between  r  and  r'  is 
given  in  (20)  and  plotted  in  Plate  II.  The  values  of  NI  are 
taken  from  Plate  I,  and  curves  of  N2  are  plotted  from  assumed 
values  of  p'  ',  r',  n,  and  d!  -r-  d.  The  reinforcement  is  seldom 
more  than  TV  d  from  the  edge,  and  the  plate  is  made  with  this 
assumption.  If  the  steel  be  nearer  than  1/10  d  from  the  edge, 
the  error  is  on  the  safe  side. 

In  case  d'  is  not  d  -r-  10  the  diagram  may  still  be  used.  In 
(21)  it  is  seen  that  d'  appears  only  in  (d  —  d'}  -f--  d,  and  so  N* 
is  readily  changed  to  suit  any  value  of  d'.  For  example,  let 
d'  be  2/10  d,  then  the  ordinate  in  Plate  II  between  the  curve  of 
p'  =  0,  and  that  of  any  other  value  of  p'  will  be  8/9  of  that 
given  in  the  diagram.  If  p'  =  1  per  cent,  and  p  =  1J  per  cent., 
the  ordinate  between  the  curves  for  p'  =  o  and  p'  =  1  per  cent. 
is  0.268  -  0.200  =  0.068  when  d'  =  1/10  d.  If  d'  =  2/10  d  the 
ordinate  will  be  8/9  X  0.068  =  0.06,  and  N2  =  0.200  -5-  0.06  = 
0.260,  instead  of  0.268  as  given  in  the  diagram. 

It  will  be  seen  that  the  factors  St,  Sc,  C,  p,  and  p'  are  not 
independent  of  each  other.  If  the  bending  moment  be  fixed 
and  the  dimensions  of  the  beam  be  assumed,  or  otherwise  deter- 
mined, the  amount  of  both  tensile  and  compressive  steel  is 
dependent  upon  the  unit  strength  assigned  to  either  the  concrete 
or  to  the  metal. 

For  example,  let  the  allowable  unit  stresses  per  square  inch 
be  Sc  =  St  =  15,000,  and  C  =  600.  Let  b  =  16  inches,  d  =  20 
inches,  p  =  2|  per  cent.,  and  p'  =  1  per  cent.  In  the  diagram 
NZ  is  0.305,  so  the  allowable  bending  moment  is,  in  pounds  per 
square  inch, 

M  =  .305  X  600  X  16  X  400  =  1,171,000. 

The  intersection  of  p  and  p'  is  on  the  vertical  through  r  and 
r',  which  in  this  case  are  15  and  12  respectively,  so  St  =  9000 
and  Sc  =  7200  pounds  per  square  inch.  These  values  are  well 
within  the  safe  limits  set  above. 

In  the  last  example,  let  it  be  required  to  determine  the  amount 


46  REINFORCED  CONCRETE 

of  tensile  steel  sufficient  to  develop  the  given  strength  of  the 
concrete  and  the  compressive  steel.  The  value  of  r  is  15,000  -f- 
600  =  25,  and  the  ordinate  through  r  =  25  intersects  p'  =  1 
per  cent,  in  p  =  1.25  per  cent.  N2  is  .255,  and  6  and  d  may 
be  varied  to  suit  the  loads.  For  d'  -r-  d  =  .1  in  (20),  r  and  r' 
are  equal  when  each  is  12.27,  and  it  is  seen  that,  for  most  com- 
binations of  p  and  p',  this  condition  is  not  realized. 

Let  it  be  required  to  find  6  and  d  when  the  length  is  20  feet, 
w  is  600  pounds  per  linear  foot,  allowable  unit  stress  in  concrete 
and  steel  500  and  16,000  Ib.  sq.  in.,  p  =  1.5  per  cent.,  and 
p'  =  1.0  per  cent.  The  weight  of  the  beam  itself  must  be  added 
to  w.  The  value  of  'N2  is  seen  to  be  0.270,  so  M  =  |  (w  +  Wi) 
202  X  12  =  .270  X  500  bd2.  If  the  weight  of  the  beam  be 
assumed,  as  nearly  as  can  be,  at  200  pounds  per  linear  foot, 
M  =  480,000  and  bd2  is  3560.  If  the  breadth  be  taken  as 
10  inches,  d  is  18.9  inches.  If  two  inches  be  added  below  the 
steel,  the  area  of  cross-section  is  209  square  inches,  and  the 
weight  per  linear  foot  is  219  pounds.  By  recomputation, 
the  depth  is  found  to  be  21  inches. 

Effect  of  Compressive  Steel.  Inspection  of  Plate  II  shows 
that  a  beam  of  certain  strength  may  be  had  by  many  different 
combinations  of  upper  and  lower  reinforcement.  For  example, 
Nz  is  0.200  when  the  percentages  are:  0,  1.5;  0.35,  1.0;  0.63, 
0.75;  and  1.2,  0.5.  The  former  combination  gives  the  higher 
stresses  in  the  compression  steel  as  compared  with  those  in 
the  tensile  reinforcement.  By  running  across  the  page  in  this 
way  on  any  given  value  of  Nz,  taking  the  sum  of  p  and  p'  at 
each  intersection,  it  will  be  seen  that  these  sums  have  a  certain 
minimum  value,  which  combination  is  the  one  to  be  chosen 
for  economy  of  steel.  In  this  example  the  sums  are:  1.5;  1.35; 
1.38,  and  1.7,  of  which,  of  course,  the  second  is  the  minimum. 
This  does  not  always  mean  that  a  beam,  so  constructed,  would 
be  cheaper  than  the  one  having  no  compressive  steel,  because 
the  cost  of  putting  in  place  two  rows  of  bars  instead  of  one  is 
apt  to  be  an  important  factor. 

The  projection  of  the  curves  of  p  and  p'  on  the  vertical  shows 
the  change  in  N2  due  to  changes  in  amounts  of  steel.  With 
no  compressive  steel  the  increase  in  N2  between  p  =  .005  and 
p  =  .025  is  .228  -  .142  =  .086  =  60  per  cent.  With  p  =  0.005 
the  increase  in  Nz  due  to  the  addition  of  2.5  per  cent,  of  com- 


ANALYSIS  OF  STRESSES 


47 


Diagram  f6r  Double  Reinf 
Rectangular  Beams 


PLATE  II. 


48  REINFORCED  CONCRETE 

pression  steel  is  .250  -  .142  =  .108  =  76  per  cent.  The  appli- 
cation of  this  principle  is  seen  in  the  following  example. 

If,  with  p  =  1.5  per  cent,  and  p'  =  0.5  per  cent.,  C  be  800, 
what  change  in  p'  is  necessary  to  reduce  this  to  600  pounds 
per  square  inch?  If  M  be  constant,  a  change  in  C  involves  a 
corresponding  change  in  N2)  which,  in  this  case,  is  to  be  25 
per  cent.  N2  for  p  =  0.015  and  p'  =  0.005  is  0.235,  which 
must  be  increased  25  per  cent,  in  order  to  decrease  C  in  like 
proportion,  or  N2  must  be  0.235  X  1.25  =  0.294.  Then  the 
value  of  p'  is  0.0145.  If  p'  =  0.005  be  unchanged  it  will  be 
necessary  to  make  p  something  beyond  the  limits  of  the  figure, 
or  about  0.035.  The  first  change  is  evidently  the  better,  as 
less  total  steel  is  thus  required. 

The  corresponding  changes  in  St  and  Sc  are  easily  found;  in 
the  first  assumption  above,  r  was  19  and  r'  was  11.6,  hence, 
St  =  800  X  19  =  15,200,  and  Sc  =  800  X  11.6  =  9280  Ibs.  per 
sq.  in.  In  the  second  case  r  and  r'  were  24  and  11.1  respectively, 
and  so  the  values  of  St  and  Sc  became  600  X  24  =  14,400  and 
600  X  11.1  =  6660  Ibs.  per  sq.  in.  respectively,  the  decrease 
being  5.2  per  cent,  and  28.5  per  cent. 

PROBLEMS 

9.  What  is  the  per  cent,  of  error  if  the  bending  moment  be  computed 
without  allowing  for  the  concrete  displaced  by  the  compression  rein- 
forcement? Assume  p  and  p'  each  1  per  cent. 

10.  A  simple  beam  has  b  =  12  inches,  d  =  20  inches,  ^   =  TV  p  =  2 

per  cent.,  p'  =  1  per  cent.,  and  working  strengths  of  steel  and  concrete 
as  15,000  and  500  pounds  per  square  inch.  What  is  the  safe  resisting 
moment  of  the  beam? 

11.  Let  the  beam  in  Problem  10  be  subjected  to  a  bending  moment 
of  400,000  Ib.  in.     Compute  the  working  stresses  in  St,  Sc,  and  C. 

12.  If,  in  Problem  10,  the  compressive  steel  be  removed,  what  increase 
in  d  will  be  necessary  to  maintain  the  strength  of  the  beam  unchanged? 

13.  What  must  be  the  value  of  p'  that  C  be  changed  to  600  Ib.  per 
square  inch  in  Problem  10? 

14.  In  Plate  II  plat  the  curves  of  p  =  .006,  .008,  and  .0175. 

15.  In  Plate  II  plot  a  curve  representing  the  sums  of  p  and  p'  for 
Nz  =  .250,  and  so  determine  the  minimum  amount  of  steel  necessary. 

Stresses  in  T-beams.  In  beams  of  rectangular  cross-section 
the  concrete  below  the  neutral  axis  serves  only  to  transfer  the 


ANALYSIS  OF  STRESSES 


49 


stress  in  the  steel  and  does  not  possess  tensile  strength.  Hence, 
considering  only  tension  and  compression,  about  half  of  the 
concrete  is  inert.  The  T-beam  is  designed  to  economize  in 
this  respect  by  concentrating  the  concrete,  where  it  is  most 
effective,  near  the  compressive  side  of  the  beam. 

If,  in  Fig.  21,  the  neutral  axis  falls  within  the  flange,  the 
analysis  of  stresses  is  the 
same  as  for  rectangular 
beams.  The  value  of  k  is 
dn  -f-  (r  -f-  n),  as  before, 
but  if  this  be  larger  than 
the  thickness  of  the  flange, 
t,  the  following  method  is 
to  be  used.  The  ratio  of 


' 

£ 

i           /« 

( 

i 

i 
/ 

i_ 

•  •  • 

^ 

FIG.  21. 


steel  cross-section  to  that  of  the  concrete  is  A  -r-  bd  =  p,  rather 
than  A  divided  by  the  actual  cross-section  of  the  beam.  The 
value  of  j  is 

3dn-2t(r  +  n) 

J     ~~    U  «    ,7™  O    *    /„       I       ™\    ^  \^/ 


6  dn  -  3  t  (r  +  n) 


As  the  sum  of  the  horizontal  stresses  is  zero, 


then 


k 


2k 
2dnA  + 


2  nA  +  2  U 
Since  the  sum  of  the  moments  is  zero, 


(25) 
(26) 


+  (6  -  6')  C 


(k-W 
3k 


(27) 


bCk  .,        ,       ,        ,.„  C(k-  <)'      ,        M 
~~  (a  -  k)  -  (b  -  b) —        (d  -  k) 


-  6')  C 


(k  - 


3  W  -  6  ktd  -f  3  dfr  -  k3  +  3  kV  -  2 


(28) 


50  REINFORCED  CONCRETE 

Substituting  the  value  of  k  =  dn  -f-  (r  +  n),  and  making  t  =  dx, 


The  first  term  is  recognized  as  being  N\  in  (16). 

This  formula  is  usually  applied  to  T-beams,  in  which  b'  is 
small  as  compared  with  6,  and  little  error  results  if  b'  be  dropped 
from  the  formula,  since  only  that  part  of  the  web  above  the 
neutral  axis  is  effective.  With  this  approximation  Plate  III 
is  constructed  from 

(30) 


=  NM2C  (31) 

The  extent  of  inaccuracy  involved  will  be  explained  below, 
page  53. 

The  outer  curve  in  Plate  III  is  the  same  as  that  for  n  —  15, 
in  Plate  I,  except  that  the  scale  is  changed.  The  curves  for 
various  values  of  t  end  in  the  outer  one  in  points  beyond  which 
values  of  p  and  r  indicate  that  the  neutral  axis  is  in  the  flange 
of  the  beam,  and  the  stresses  are  the  same  as  though  the  section 
were  rectangular. 

Let  the  T-beam  have:  6  =  36  inches;  d  =  20  inches;  t  =  4 
inches;  b'  =  10  inches,  working  strengths  of  concrete  and  steel 
600  and  15,000  Ib.  sq.  in.,  and  let  the  reinforcement  be  six 
f-inch  bars.  It  is  required  to  determine  the  safe  resisting  mo- 
ment of  the  beam.  The  percentage  of  reinforcement  is  6  X  9/16 
-=-  36  X  20  X  100  =  .47,  and,  from  the  diagram,  N3  =  0.125 
and  r  —  37.  With  this  value  of  r  the  full  strength  of  the  con- 
crete cannot  be  utilized,  as  600  X  37  far  exceeds  the  safe  stress 
in  St,  so  the  working  stress  in  the  concrete  is  15,000  +  37  =  405 
Ib.  sq.  in.  Then  M  =  0.125  X  36  X  400  X  405  =  729,000  Ib-in. 

The  failure  of  T-beams  is  usually  due  to  lack  of  sufficient 
steel  rather  than  to  compression  of  the  concrete.  Very  often 
the  flange  is  a  part  of  a  slab  forming  a  floor,  and  it  is  impossible 
to  know  how  wide  to  assume  the  beam.  In  the  above  example, 
if  b  =  22.3  inches,  p  is  .76  per  cent,  and  r  is  25,  as  the  allowable 
stresses  indicate.  According  to  adopted  practice  (see  page  148) 


ANALYSIS  OF  STRESSES 


51 


017         016        0.15         O.U         0.13         013         0. 
Values  of  Nlm~M-N0  C  M 


PLATE  III. 


52 


REINFORCED  CONCRETE 


the  flange  width  might  be  as  much  as  42  inches.  With  such 
an  assumed  width  the  compressive  strength  of  the  concrete  is 
seldom  developed. 

Tests  of  Reinforced  Concrete  T-beams.  The  tests  given  in 
the  following  table  have  been  made  recently  and  are  by  the 
most  reliable  investigators. 

TABLE  OF  TESTS  OF  T-BEAMS 


No. 

Mix- 
ture 

a  1 

<& 

Hori- 
zontal 
Rein- 
force- 

Web 
Reinforcement 
Rods 

Si 
sJ 

J2  % 

<uj3 

*3 

Load 
Lbs. 

Stress.    Lbs. 
Per  Sq.  In. 

Shearing  Stress  in 
Web,  b'  X  d. 

Lbs.l 
Per 

Kind  of  Fail- 

ment 

% 

t 

b 

I/ 

d 

C 

[S 

S: 

ure 

7i 

1-2-4 

60 

1.05 

10-i*  stirrups 

-! 

1C, 

S 

0 

46,700 

64,300 

293 

Tension 

I4 

60 

1.10 

Spaced  6* 

1C, 

s 

0 

32,410 

— 

41,500 

203 

I* 

" 

60 

1.10 

apart  at 

Hi 

S 

10 

30,100 

— 

38,100 

188 

M 

ll 

«i 

60 

0.93 

either  end. 

24 

s 

10 

55,700 

— 

57,500 

347 

" 

/6 

" 

60 

0.92 

In  Nos.  6,  8, 

24 

s 

0 

39,300 

— 

40,700 

246 

" 

/« 

•' 

60 

0.92 

5,  2-1"  rods 

24 

s 

10 

40,100 

— 

41,200 

2oO 

'* 

h 

«« 

60 

1.05 

were  bent  up 

52 

s 

0 

80,500 



55,700 

503 

M 

h 

«• 

60 

1.05 

In  No.  9,  3-2" 

tt 

s 

10 

83,300 

— 

57,400 

521 

" 

ll 

M 

60 

0.97 

plain  rods 

1 

52 

8 

10 

50,900 

— 

37,600 

318 

" 

were  bent  up. 

Mi 

•« 

30 

0.74 

None 

3 

24 

8 

}\ 

22,000 

1510 

34,500 

— 

Tension 

Mi 

«• 

30 

0.75 

" 

a 

24 

x 

»' 

22,000 

1495 

33,300 

— 

M-A 

" 

30 

0.83 

" 

a 

21 

s 

)" 

22,000 

1410';  38,100 

— 

" 

M4 

•• 

30 

0.91 

" 

3 

24 

s 

t. 

24,000 

1570  21,600 

— 

Compression 

Ml 

" 

30 

1.04 

" 

3 

24 

s 

).. 

28,000 

1780 

29,600 

— 

Wi 

1-2-4 

30 

0.94 

Rods  bent  up 

2 

g 

3 

4; 

4,000 





107 

Bond 

Wt 

" 

30 

0.94 

11        «      it 

2 

g 

3 

I 

5,100 

— 

— 

136 

•• 

W3 

M 

30 

0.521 

a 

9 

3 

I 

4,750 



— 

127 

M 

W4 

" 

30 

0.52 

2 

9 

3 

I 

4,000 

— 

— 

107 

«« 

W5 

'« 

30 

0.52 

Rods  and  \" 

2 

9 

3 

H 

9,380 

— 

— 

238 

i     Tension  and 

W6 
W7 

M 

30 
30 

0.52 
1.05 

stirrups  spaced 
3i"  apart. 
In  Nos.  5,  6. 

2 

a 

9 

g 

3 
3 

t] 

4J 

9,400 
12,850 



I 

246 
330 

'         shear. 
I     Compression 
)          and  shear. 

w* 

" 

30 

1.05 

7,  and  8  the 

a 

(i 

3 

4j 

13,550 

— 

— 

349 

Compression 

W* 

1 

30 

0.39 

stirrups  were 

a 

12 

3 

1? 

8,400 

— 

— 

216 

Shear 

Win 

' 

30 

0.39 

ft* 

a 

12 

3 

t-i 

8,000 

— 

— 

205 

" 

Wn 

' 

30 

0.78 

2 

12 

3 

4 

11,400 

— 

— 

304 

« 

Wl2 

1 

30 

0.78 

2 

12 

3 

4 

12,700 





346 

«« 

Wn 

' 

30 

0.52 

Expanded 

2 

1 

3 

4 

7,800 

— 

— 

217 

«« 

Wl4 

30 

0.52 

Metal 

2 

I 

3 

4 

7,000 

— 

— 

190 

t* 

Wl5 

« 

30 

1.05 

a 

1 

3 

4 

13,800 

—  , 

— 

384 

Tension  and 

shear 

w* 

30 

1.05 

2 

1 

3 

4 

12,600 

350 

Compression 

The  tests  numbered  7  are  from  Bulletin  No.  12,  University  of 
Illinois  Engineering  Experiment  Station,  1907.  Those  num- 
bered M  were  made  by  Professor  F.  B.  McKibben  at  the  Massa- 
chusetts Institute  of  Technology.  Those  numbered  W  are  from 
Bulletin  No.  1,  Vol.  4,  University  of  Wisconsin,  1907. 

The  following  are  additional  data  concerning  these  tests.  In 
the  beams  marked  /  the  concrete  had  a  compressive  strength  .in 
cubes  of  1820  Ibs.  per  sq.  in.  In  Nos.  1,  3,  2,  and  5,  the  bars  were 


ANALYSIS  OF  STRESSES  53 

deformed  and  had  a  yield  point  of  53,800  Ibs.  per  sq.  in.  The 
other  steel  was  plain  with  yield  point  of  38,300  Ibs.  per  sq.  in. 
The  stirrups  were  of  the  deformed  steel.  The  span  was  10  feet. 
The  loads  in  all  beams  were  at  the  third  points.  In  the  beams 
marked  M  the  concrete,  in  cubes,  had  an  average  compressive 
strength  of  1790  Ibs.  per  sq.  in.  The  span  was  12  feet.  In 
the  beams  marked  W  the  span  of  the  first  four  was  6  feet.  In 
the  others  this  was  reduced  to  5  feet,  while  the  whole  length 
of  the  beam  was  6f  feet.  The  strength  of  compression  cubes 
of  this  concrete  was  1120  Ibs.  per  sq.  in.  Two  of  the  three 
rods  were  bent  up  in  beams  1-6,  9,  10,  13,  and  14,  while  four  of 
the  six  were  bent  up  in  other  cases.  The  percentage  of  steel  is 
in  every  case  based  upon  the  area  bd. 

As  is  usual  in  such  tests,  there  were  many  failures  by  tension 
in  the  steel. 

Influence  of  Web  Compression.  In  order  to  show  the  addi- 
tional bending  moment,  due  to  the  part  of  the  web  that  lies 
above  the  neutral  axis,  an  example  will  be  chosen  in  which  the 
assumed  dimensions  are  not  in  accordance  with  usual  practice. 
This  difference  between  the  true  and  the  approximate  bending 
moments  is  given  by  subtracting  (30)  from  (29),  or, 

, , ,  „  /3  r  +  2  n  ,  r  +  2  n         ,  r  +  n\ 

V&  C    ai\ — N?  n  —  x  +  x2 — 5 x3  —^ — 

\6  (r  +  n)2  2n  3  n  ) 

should  be  added  to  values  of  M,  as  taken  from  Plate  III.  It 
will  be  noted  that  the 'first  fraction  within  the  parentheses  is 
given  by  the  outer  curve  in  Plate  III,  and  that  the  remainder 
of  the  expression  is  shown  by  the  curves  for  various  values  of  t. 

Hence,  the  correction  to  be  applied  for  web  compression  is 
found  by  multiplying  the  ordinate  intercepted  between  the  outer 
curve,  and  that  for  the  given  value  of  t,  in  Plate  III,  by  b'd2C. 

For  example,  let  6  =  40  inches,  d  =  32  inches,  &'  =  18  inches, 
p  =  0.005,  t  =  4  inches,  C  =  500,  and  St  =  16,000  Ib.  sq.  in. 
Since  t  =  d  -T-  8,  the  intercept  is,  on  the  ordinate  for  p  =  0.005, 
0.143  -  0.095  =  .048.  Then,  .048  X  18  X  32  X  32  X  16,000 
-T-  32  =  442,000  Ib.  in.  By  (27),  M  =  .095  X  32  X  32  X  40  X 
16,000  4-  32  =  1,950,000  Ib.  in.,  and  the  whole  moment  is 
1,950,000  +  442,000  =  2,392,000  Ib.  in.,  or  the  web  adds  22 
per  cent,  to  the  moment  as  usually  computed. 


54  REINFORCED  CONCRETE 

To  show  the  effect  of  the  web  compression  in  a  T-beam  of 
ordinary  design,  let  bf  =  12  inches,  c  =  6  inches,  and  the  beam 
otherwise  like  the  last.  Then  the  correction  for  web  compres- 
sion is  (.143  -  .123)  12  X  322  X  500  =  123,000  Ib.  in.  in  a  total 
of  2,373,000  Ib.  in.,  or  about  5.2  per  cent.  The  error  is  seen  to 
increase  rapidly  as  the  slab  or  flange  becomes  thinner  and  the 
per  cent,  of  steel  increases.  Thus,  if  the  depth  be  32  inches 
and  the  percentage  of  steel  be  0.5,  the  error,  for  t  —  4  inches, 
is  2.4  times  as  much  as  when  t  =  6  inches,  if  the  width  of  the 
web  be  unchanged. 

Depth  of  T-beams.  Let  it  be  required  to  find  depth  of  a 
T-beam  to  have  a  resisting  moment  of  1,500,000  Ib.  in.  The 
flange  is  40  inches  wide  and  5  inches  deep.  The  width  of  the 
web  is  determined  by  the  number  and  size  of  the  reinforcing 
bars,  and  the  necessary  spaces  between  and  outside  them. 
According  to  the  recommended  practice  (see  page  150),  b'  must 
be  1.5  +  2J  n  diameters  of  the  bars,  where  n  is  the  number 
of  such  bars.  Since  the  curves  for  t  in  Plate  III  depend  upon 
values  of  d,  it  is  necessary  to  assume  the  depth  and  make  the 
value  of  p  to  correspond.  Let  St  and  C  be  assumed  not  to 
exceed  16,000  and  600  Ibs.  sq.  in.  respectively,  and  let  d  be 
taken  as  24  inches,  for  trial.  Then,  as  r  is  16,000  -f-  600,  p  is 
.69  per  cent,  and  t  =  .208  d,  so  M  =  .135  X  40  X  24  X  24  X 
600  =  1,870,000  Ib.  in.  The  width  of  the  web  is  found  when 
the  number  and  size  of  the  rods  have  been  decided  upon.  Since 
p  is  0.69  per  cent.,  A  =  24  X  40  X  .0069  =  6.6  square  inches, 
and  7  one-inch  square  bars  will  be  sufficient.  The  width  is 
b'  =  1.5  +  2.5  X  7  =  19  inches,  and  the  weight  of  the  beam 
must  be  found  and  the  dead  load  moment  added  to  1,500,000. 
If  the  result  be  less  than  the  computed  resisting  moment  the 
design  is  safe.  If  the  length  be  20  feet  the  total  moment  is 
found  to  be  1,875,000  Ib.  in.  As  b'  is  rather  large,  it  will  be 
better  to  assume  a  greater  depth  and  make  the  computations 
again;  or,  a  number  of  smaller  bars  to  provide  sufficient  area 
may  be  inserted  in  two  rows.  If  p  be  0.6  per  cent,  and  d  be 
26  inches,  8  one-inch  diameter  rods  will  be  sufficient,  and  may 
be  placed  in  two  rows,  the  width  of  web  will  be  14.0  inches,  and. 
the  resisting  moment  will  be  1,900,000  Ib.  in.,  while  the  moment 
of  loads  is  found  to  be  1,833,000  Ib.  in. 


ANALYSIS  OF  STRESSES 


55 


PROBLEMS 

16.  A  T-beam  has  dimensions  as  follows:  b  =  42  inches;  b'  —  12  inches; 
t  =  6  inches;  d  =  20  inches.     The  reinforcement  consists  of  six  three- 
quarter-inch  rods,  the  safe  stresses  for  St  and  C  are  16,000  and  500  Ib. 
per  sq.  in.  respectively,  and  n  is  15.     What  is  the  safe  resisting  moment 
of  the  beam? 

17.  In  the  last  problem,  for  what  thickness  of  flange  will  the  neutral 
axis  fall  a  distance  t  from  the  top? 

18.  In  Problem  16  what  per  cent,  of  the  whole  bending  moment  is 
due  to  the  web  of  the  beam? 

19.  In  Plate  III  how  are  the  vertical  lines  indicating  percentages  of 
steel  located? 

20.  A  beam  20  feet  long  is  to  carry  a  uniform  load,  besides  its  own 
weight,  of  500  pounds  per  linear  foot,  and  a  concentrated  load  of  4000 
pounds  at  the  middle.     The  effective  width  of  the  flange  is  48  inches 
and  its  depth  is  5  inches.     If  the  depth  of  beam  below  the  flange,  or 
slab,  be  limited  to  15  inches,  what  amount  of  steel  is  required? 

21.  In  the  last  problem  what  saving  in  per  cent,  of  steel  is  possible 
if  the  effective  depth  be  increased  25  per  cent.? 

22.  If,  in  Problem  20,  the  depth  be  unlimited,  compute  its  value  so 
that  St  =  16,000  and  C  =  400  Ib.  per  sq.  in. 

23.  Deduce  the  value  of  k  for  a  T-beam,  in  terms  of  n,  d,  and  r.    Why 
is  this  value  the  same  as  in  a  beam  of  rectangular-  section? 

Analysis  of  Flexural  Stresses  under  Ultimate  Loads.  The 
foregoing  flexural  formulas  have  been  based  upon  the  assumption 
that  the  stresses  and  corresponding  deformations  are  in  a  con- 
stant ratio.  While  this  theory  is  practically  true  for  working 
loads,  it  is  not  even  approximately  so  for  loads  that  are  nearly 
the  ultimate  for  the  beam.  When  stresses  and  deformations 
are  plotted  as  abscis- 
sas and  ordinates,  it 
is  noted  that  the 
diagram,  called  the 
"stress  =  strain  dia- 
gram," is  a  curve 
that  follows  very 
closely  a  parabola, 
having  its  origin  at 
the  point  indicating 
the  maximum  stress  and  elongation,  and  its  axis  horizontal. 
Fig.  22  shows  two  conceptions  of  the  parabolic  stress  diagram. 
In  (a)  the  axis  is  on  the  upper  line,  and  C  is  the  assumed  or 


§/*» 


56 


REINFORCED  CONCRETE 


observed  stress  in  the  concrete.  In  (b)  the  axis  is  on  C",  which 
is  the  ultimate  stress,  and  C  is  any  other  stress  that  may  be 
considered.  In  (6)  the  value  of  C'  may  be  computed  from  any 
observed  stress,  C  and  corresponding  deformation  c. 

These  analyses  do  not  hold  if  the  steel  be  stressed  beyond 
its  elastic  limit.  In  other  words,  the  beam  when  loaded  to 
destruction,  should  not  break  through  failure  of  the  steel. 

Stresses  under  Ultimate  Loads.  In  Fig.  23  the  shaded  part 
is  a  full  parabola  having  the  axis  in  C.  The  equation  of  this 

curve  is  y2  =  mx,  and  it  may  be 
shown  that  ab  =  bg  when  go  is  a 
tangent  to  the  parabola  at  o. 
The  area  of  abo  is  f  ab.ao,  and 
the  center  of  gravity  of  abo  is  f 
ao  =  |  k  above  o.  Since  the 
tangent  of  the  angle  between  the 
vertical  and  the  line  tangent  to 
the  curve  at  o  is,  at  o,  the  ratio  of 
the  unit  stress  to  the  unit  defor- 
mation, 


FIG.  23. 


tan  aog  =  tan 


Ec 


in  which  Ec  is  the  initial  coefficient  of  elasticity  of  the  concrete. 
(See  page  21.) 
For  equilibrium  the  sum  of  the  horizontal  forces  is  zero,  so 


f  bCk  =  ASt  =  pbdSt 
The  resisting  moment  equals  the  bending  moment  and 

or  f  bCkd  -  \  bCk2  =  M 

As  sections  remain  plane  during  bending 

—:(d-k)  =2C:k 
n 

k  = 


(32) 


(33) 


St  +  2  Cn 

2dn 
r+2n 


(34) 


Also  from  (32)  and  (34)       k  =  dV3  pn  +  ( J  pn)*  -  |  pnd    (35) 


ANALYSIS  OF  STRESSES  57 

=  I  pdr  (36) 

From  (34)  and  (36)          p  =  3  (r  +*  n)  ,  (37) 

Putting  (34)  in  (33),      M  =  bd*C  ^  j"  \  "   n  (38) 

o  ^r  ~f~  Zn)2 


(39) 
This  may  be  used  to  find  St  by  substituting  St  -r-  r  for  C. 

Values  of  N4  are  plotted  in  Plate  IV  as  ordinates  for  n  =  10, 
15,  and  20,  and  percentages  of  reinforcement  are  fixed  in  accord- 
ance with  (37).  The  use  of  this  plate  is  similar  to  that  of  Plate  I, 
and  the  same  limitations  exist  as  to  what  may  be  assumed. 

For  example,  what  is  the  ultimate  resisting  moment  of  a 
concrete  beam  in  which  b  =  10  inches,  d  =  18  inches,  St  =  elas- 
tic limit  of  the  steel  =  50,000  Ib.  per  sq.  in.,  C  =  ultimate 
strength  of  concrete  =  2000  Ib.  sq.  in.,  and  the  reinforcement 
consists  of  four  f-inch  diameter  steel  rods?  The  percentage  of 
steel  is  6  X  &  X  .7854  -5-  10  X  18  =  1.46.  (See  page  156  for 
data  concerning  steel  rods  and  bars.)  From  the  diagram  r  is 
25,  and  N4is  .29,  so  M  =  .29  X  10  X  18  X  18  X  2000  =  1,870,- 
000  Ib.  in.  If  the  percentage  of  steel  were  less  than  that  given, 
as,  say,  1.2,  r  would  be  28.4  and  C  would  equal  50,000  -r-  28.4  = 
1760  Ib.  sq.  in.,  and  the  beam  would  fail  in  the  steel.  In  such 
cases  the  formulas  and  Plate  IV  do  not  apply  as  they  are  made 
with  the  assumption  that  the  steel  is  not  stressed  beyond  its 
elastic  limit.  For  the  solution  of  such  cases  see  page  62. 

As  an  example  of  design  let  it  be  required  to  find  the  dimen- 
sions of  a  beam  20  feet  long  to  carry  a  uniform  load  of  600  pounds 
per  linear  foot,  with  a  factor  of  safety  of  4,  or  an  ultimate  load 
of  2400  pounds  per  foot  of  length.  Let  the  elastic  limit  of 
steel  be  50,000,  and  the  ultimate  compressive  strength  of  con- 
crete 1800  Ib.  sq.  in.  If  these  stresses  be  reached  at  the  same 
time,  r  is  27.8,  and  p  must  be  .0125.  2V4  is  .278.  Then  omitting, 
for  the  time,  the  weight  of  the  beam,  M  =  J  X  2400  X  20  X 
20  X  12  =  1,440,000  Ib.  in.  =  .278  X  1800  X  fed2.  From  this 
bd2  =  2880.  The  breadth  may  be  assumed,  as  a  trial,  at  10 
inches,  then  d  becomes  17  inches.  There  must  be  2  inches  of 
concrete  below  the  steel,  so  the  cross-section  of  the  beam  will 
be  10  X  19  =  190  sq.  in.,  and  this  amount,  at  150  pounds  per 


58 


REINFORCED  CONCRETE 


PLATE  IV. 


ANALYSIS  OF  STRESSES 


59 


cubic  foot,  adds  about  33  per  cent,  to  the  bending  moment, 
so  bd2  should  be  2880  X  1.33  =  3830,  and  b  and  d  may  be  12 
and  18  inches  respectively.  The  reinforcing  area  is  12  X  18  X 
.0125  =  2.7  sq.  in.,  and  6  —  f  inches  rods  will  be  sufficient. 

By  comparing  Plate  I  with  Plate  IV  it  is  noted  that  the  straight 
line  relation  calls  for  only  about  half  as  large  percentage  of  steel 
to  correspond  with  any  given  r  as  does  the  parabolic  relation. 
On  the  other  hand,  with  the  same  amount  of  steel,  a  greater 
moment  is  indicated  in  Plate  IV,  and  the  straight  line  formulas 
are  the  more  conservative. 

PROBLEMS 

24.  Design  a  beam  of  18  feet  span  to  safely  carry  a  uniform  load  of 
4  times  its  own  weight,  using  1:2:4  concrete  and  mild  steel. 

25.  A  beam  24  feet  long,  10  inches  wide,  and  16  inches  effective  depth 
is  reinforced  with  6  —  |  inch  square  bars.    It  breaks  by  compression  under 
a  uniform  load,  including  its  own  weight,  of  1650  pounds  per  linear  foot. 
What  is  the  probable  stress  in  the  concrete  if  the  steel  be  capable  of 
an  elastic  limit  of  42,000  Ibs.  per  in.? 

26.  Solve  the  last  problem,  using  both  n  =  10  and  n  =  20.     Explain 
the  meaning  of  the  difference  in  results. 

27.  Use  a  factor  of  safety  of  4,  and  solve  Problem  24  by  the  straight 
line  relations  and  Plate  I. 

The  General  Parabolic  Stress-Strain  Relation.  This  analysis 
is  useful  in  interpreting  the  results  of  tests  on  beams  in  which 

the  stresses  range  from 
zero  to  any  value  up 
to  the  ultimate.  The 
formulas  were  devel- 
oped by  Professor  A. 
N.  Talbot,  and  pub- 
lished in  1905  in  Bul- 
letin No.  4  of  the 
University  of  Illinois 
Engineering  E  x  p  e  r  i- 
ment  Station. 

Fig.    24    shows    the 
stress-strain      diagram 
in  which  C'  is  the  ulti- 
mate unit  stress  and  c' 
is  the  ultimate  deformation  of  the  concrete.     Any  other  stress 


FIG.  24. 


60  REINFORCED  CONCRETE 

and  corresponding  deformation  are  shown  as  C  and  c'.  The 
tangent  to  the  parabola  at  o  indicates  the  initial  coefficient  of 
elasticity  of  the  concrete,  and  tan  aob  =  tan  oc  =  2  C'  -r-  c'  =  Ec. 
The  origin  of  the  parabola  is  at  /'  and  f'b'  is  the  axis.  In  the 
equation  for  the  parabola,  x  =  |  c'Ec,  when  y2  =  c'2,  so  it  may 
be  written 

2  c'x 

y*  =  —ET 


(40) 
(41) 


From  Fig.  24 

(c' 

*i% 

V2«  ~  C 

9«t<-5 

and 

C 

EcC 

-I  -q/2 

also 

C 

C' 

ECC  (1  - 

<7/2)       2 

5  (1  -  g/2) 

t  Ecc' 

The  area  obf 

=  C'c-\c 

'C'  +  J  0 

c'  -  c)  (C"  - 

The  area  oab  =  oa'6'4  =  C'c'q2  (43) 


=  f  C'c  -  J  Cc'  +  i  Cc  (42) 

=  oa'6'^2  =  C'c'q2 

Area  of  parabola  _  $q  —  C/C'  Q  —  ^  q)  ,     . 

Area  of  triangle  q2 

=  1  ~  1  q  (45) 

This  expression  shows  the  relation  between  the  sums  of  the 
compressive  stresses  in  the  beam  according  to  the  parabolic, 
and  the  straight  line  theories  when  k  is  substituted  for  c. 

In  equation  (40)  C  is  the  stress  in  the  outer  fiber  according  to 
the  parabolic,  and  Ecc  is  the  stress  at  that  place  according  to 
the  straight  line  assumption,  or  ab  in  Fig.  24. 

The  center  of  gravity  of  ob'f  is  f  ob'  below  b'.  As  the  area 
of  ob'f  is  f  06'  X  b'f,  and  the  area  of  obf  has  been  found  (42), 
the  distance  of  the  center  of  gravity  of  obf  below  b  can  be  found 
in  the  ordinary  way.  Letting  c  equal  k 

i  =il^b  (46) 

For  equilibrium,  the  horizontal  forces  are  zero,  so 

b  (f  C'k  -  i  C  -  +  i  Cfc)  =  ASt  =  pbdSt  (47) 


ANALYSIS  OF  STRESSES  61 


from  which  bCk    *  ~  q   x  =  ASt  =  pbdSt  (48) 

&  (^  —  q) 

As  the  bending  moment  equals  the  resisting  moment, 

M  =  bCk     3  ~  q     (k  -  z)  +  ASt  (d  -  k)  (49) 

6  (4  —  Q) 

4  (3  d  -  k)  -  q  (4  d  -  fc) 


Since  a  vertical  section  remains  plane  during  bending,  from 
(37)  and  from  Fig.  24. 

St      2c 


=  (d  -  k) :  k 

/t      A  —  y 

2dn 
and  k  =    , 

r  (2  —  q)  +  2  n 

a  2n 


From  (48)  and  (45) 


P   ~ 


(2  --f  2  n 
(3  ~  9)  ft2 


Q(d-k)nd 


Substituting  (48)  in  (47) 

M  -  bd*  C  2(3-g)(r(2-g)+2n)-(4-g)n 
3  (2  -  q)  (r  (2  -  g)  +  2  n)2 

=  AW2C  (55) 

If  q  be  zero,  the  above  formulas  become  the  same  as  in  the 
straight  line  theory,  and  if  q  be  made  equal  to  unity,  these  formu- 
las reduce  to  the  corresponding  ones  under  the  full  parabolic 
assumption,  as  of  course  they  should. 

In  order  to  plot  JVs,  the  fraction  in  (53)  may  have  q  succes- 
sively J,  ^,  |,  or  any  other  assumed  value,  and  a  curve  is  made 
for  each  of  these  assumptions.  For  use  with  safe  working  loads 
and  stresses  q  may  be  taken  as  J.  The  larger  values  of  q  are 
useful  in  investigating  stresses  near  the  ultimate. 

In  Plate  V  the  q  curves  are  plotted,  with  values  of  N$  as 
ordinates  and  values  of  r  as  abscissas,  from  equation  (54)  ;  while 
the  curves  for  percentage  of  steel  are  taken  from  equation  (53). 
To  illustrate  the  use  of  this  diagram,  let  it  be  required  to  find 


62  REINFORCED  CONCRETE 

the  stresses  in  the  steel  and  concrete  in  a  beam  having  the 
following  dimensions.  Breadth,  8  inches;  depth,  11  inches, 
total,  and  10  inches  effective;  span  12  feet,  and  2.76  per  cent, 
of  steel  reinforcement.  A  total  load  of  15,000  pounds  was 
applied  at  the  one-third  points,  causing  rupture  by  compression 
of  the  concrete.  The  bending  moment  is  7500  X  4  X  12  = 
360,000  Ib.  in.  In  the  diagram,  with  q  =  1  and  p  =  2.76  per 
cent.,  Nb  is  found  to  be  0.33  and  r  is  15.6.  Then  0.33  C  X  8  X 
10  X  10  =  360,000  and  C  =  1363  and  St  =  21,300  Ib.  sq.  in. 
This  beam  was  nominally  a  1:3:6  mixture,  and  must  have 
been  only  fairly  well  made  to  give  this  result  for  ultimate  com- 
pression. The  working  load  for  this  beam  is  found  by  taking 
q  =  \  in  the  diagram.  Then  .248  X  600  X  8  X  10  X  10  =M 
=  119,000  Ib.  in.,  and  the  load  is  4960  pounds,  giving  a  factor 
of  safety  of  about  3.  Using  the  straight  line  formulas  and 
Plate  I,  Ni  is  0.236,  so  M  is  113,300  Ib.  in.,  and  the  computed 
load  4720  pounds. 

Let  it  be  required  to  design  a  beam  to  carry  a  load  which 
makes  the  bending  moment  500,000  Ib.  in.  If  the  working 
stresses  be  desired,  q  is  taken  as  J,  C  =  500,  and  St  =  16,000 
Ib.  per  sq.  in.  From  the  diagram  JV5  is  0.161  and  p  is  about 
0.6  per  cent.,  so  0.161  X  500  bd2  =  500,000  and  bd2  =  6210. 
If  b  =  12  inches  the  effective  depth  will  be  23  inches,  and  the 
reinforcement  may  be  5  round  rods,  f  inch  diameter,  or  4  round 
rods,  f  inch  diameter.  (See  page  156.)  It  will  be  noted  that 
the  assumption  of  q  =  J  gives  a  factor  of  safety  of  16  -r-  7  =  2.3. 

In  the  following  example  the  reinforcement  is  not  sufficient 
to  develop  the  full  strength  of  the  concrete  and  failure  would 
take  place  first  in  the  steel.  Let  S't  =  50,000  and  C'  =  2000 
Ibs.  per  sq.  in.,  b  =  10  inches,  effective  depth  =  18  inches  and 
p  =  0.012.  It  is  required  to  find  the  ultimate  bending  moment 
for  the  beam.  With  r  =  25  and  p  =  1.2  per  cent,  in  Plate  V, 
q  is  seen  to  be  a  little  more  than  f ,  in  which  case  C  =  0.94  X 
2000  =  1880  Ib.  per  sq.  in.  (see  small  diagram  a  on  Plate  V). 
With  this  value  of  C  and  r  =  25,  the  strength  of  the  steel  would 
not  be  the  ultimate,  and  the  proper  value  of  q  is  sought  by  trial. 
If  r  be  assumed  at  26,  C  is  about  19,500  and  Stf  is  too  large. 
If  r  be  taken  as  25.6,  q  is  .85,  N*>  is  .252,  C  is  2000  X  .97,  and 
St'  is  49,700  Ib.  per  sq.  in.  Then  M  =  .252  X  1940  X  10  X  18 
X  18  =  1,585,000  Ib.  in. 


ANALYSIS  OF  STRESSES 


63 


Diagram  for  General  Parabolic 


PLATE  V. 


64  REINFORCED  CONCRETE 

Let  a  beam  be  subjected  to  a  bending  moment  of  350,000 
Ib.  in.  If  b  =  10  inches,  effective  depth  =  17  inches,  and 
p  =  1  per  cent.,  what  will  be  the  stresses  in  the  steel  and  in  the 
concrete?  Here  q  and  the  value  of  C  are  dependent  upon  each 
other  and  are  both  unknown.  It  may  be  assumed  that  the 
stress  will  be  a  fourth  of  the  ultimate,  in  which  case  (a)  of  Plate 
V  gives  q  as  about  .15.  Then  N$  is  .186  and  r  is  22,  for  p  =  1 
per  cent.  So  0.186  C  X  10  X  17  X  17  =  350,000  and  C  =  650. 
St  =  650  X  22  =  14,300  Ib.  per  sq.  in.  If  the  concrete  be  such 
that  the  ultimate  strength  is  very  different  from  650  X  4  another 
assumption  must  be  .made  and  the  computations  repeated. 

If  the  straight  line  relation  be  used,  q  is  zero,  N&  is  0.179  and 
C  and  St  become  675  and  14,400  Ib.  per  sq.  in.  respectively. 
For  working  loads  and  stresses  this  method  of  computation  is 
safe  without  being  wasteful  of  material. 

When  a  beam  is  tested  with  known  loads,  it  is  usual  to  deter- 
mine the  value  of  k  by  extensometers  and  direct  measurements. 
If  k  be  substituted  in  (43)  z  becomes  known  in  terms  of  the  still 
unknown  q.  The  value  of  z  varies  within  rather  narrow  limits: 

q  =  0,  z  =  J  k;  q  =  i,  z  =  if  fc;  q  =  \,  z  =  •&  k;  q  =  f , 

z  =  Iffc;  q  =  1,  z  =  f  k. 

So  if  k  be  known,  z  is  found  without  great  error  even  if  q  be 
only  approximately  assumed.  The  manner  in  which  the  beam 
fails  is  a  fair  indication  of  q.  If  the  failure  be  by  crushing  the 
outer  fibers  of  the  concrete,  q  is  unity;  if  the  steel  break,  q  is 
probably  about  |,  and  failure  by  diagonal  tension  usually  in- 
dicates a  stress  in  the  steel  due  to  q  =  J  to  J.  By  means  of 
the  extensometers  on  the  beam  and  on  direct  compression 
pieces  of  the  same  kind  of  concrete,  q  is  determined  with  greater 
accuracy. 

When  z  is  known  the  resisting  moment  of  the  beam  is  often 
most  conveniently  stated  in  terms  of  the  steel,  or 

pbd  St  (d  -  z)  =  M 

where  M  is  the  moment  of  the  loads. 

The  diagram  in  Plate  VII  shows  values  of  q  in  terms  of  k 
and  p  for  n  =  15.  It  is  readily  seen  that,  for  a  given  percentage 
of  steel,  k  must  be  found  between  the  upper  and  lower  curves 
intercepted  by  that  ordinate.  The  measured  value  of  k  often 


ANALYSIS  OF  STRESSES 


65 


PLATE  VI. 


66 


REINFORCED  CONCRETE 


falls  outside  these  limits,  indicating  that  n  is  other  than  15, 
upon  which  assumption  Plate  VII  is  made.  In  the  following  table 
k  is  taken  from  measurement,  6  =  8  inches,  and  the  effective 


Per  Cent 
of  Steel 

» 

Moment  of 
Loads 
Lb.-In. 

q 

Pounds  per  Sq.  In. 

Manner  of  Failure 

C 

Sc 

.74 

.380  d 

192,000 

I 

1362 

36,800 

Tension  in  steel. 

.74 

.410  d 

168,000 

I 

1100 

33,000 

«         «      « 

1.23 

.470  d 

288,000 

i 

4 

1762 

35,100 

ii        d      ii 

1.60 

.501  d 

312,000 

1 
4 

1775 

29,300 

Diagonal  tension. 

1.66 

.505  d 

336,000 

i 

1900 

30,400 

«              tt 

1.84 

.552  d 

336,000 

i 

1720 

28,400 

(I                           U 

2.21 

.605  d 

360,000 

3 

4 

1065 

26,100 

Compression   follow- 

ing  diagonal  tension. 

depth  is  10  inches.  The  method  of  computing  C  and  St  is 
shown  by  taking  the  sixth  beam  as  an  example.  In  Plate  VII 
for  k  and  p  as  given,  q  is  found  to  be  slightly  more  than  ^.  Then 
in  Plate  V,  with  q  =  i  and  p  =  0.0184,  N  is  0.244  and  r  is  16.5, 
then 

0.244  C  X  8  X  10  X  10  =  336,000 
C  =  1720 
St  =  170  X  16.5  =  28,400 

The  stress  in  the  steel  may  be  found  otherwise,  as  indicated 
above,  by  first  finding  z  from  k  when  q  =  J. 

d  -  z    =  10  -  ft  X  5.52  =  8.06  inches 
.0184  X  St  X  10  X  8  X  8.06  =  336,000 
St  =  28,325  Ib.  per  sq.  in. 

which  result  agrees  substantially  with  that  obtained  from  the 
Plate  V. 

To  illustrate  the  effect  of  the  assumption  as  to  q,  let  the  first 
beam  in  the  above  table  be  investigated  for  q  =  0,  q  =  J,  q  =  i, 
9  =  |,  and  q  =  1.  The  corresponding  values  of  St  are  readily 
found  to  be:  36,500;  36,800;  37,200;  37,400,  and  38,200  pounds 
per  square  inch,  and  the  difference  between  the  highest  and 
the  lowest  is  but  about  6  per  cent,  of  the  former.  If  q  =  0, 
and  q  =  1  be  omitted,  the  results  agree  within  1  per  cent,  whether 


ANALYSIS  OF  STRESSES 


67 


fc  4-  d  in  terriaS  of  q  and 
percentage  of  steel 


PLATE  VII. 


68  REINFORCED  CONCRETE 

q  be  assumed  as  |,  J,  or  J.  The  computed  stresses  in  the  con- 
crete, however,  vary  materially  according  to  the  assumption  as 
to  q. 

Plate  VI  contains  diagrams  for  computing  N$  in  (55)  when 
n  is  either  12  or  18.  The  next  to  the  last  of  the  above  beams 
gives  values  of  C:  1820,  1720,  and  1630,  and  for  St:  28,200, 
28,400,  and  28,600  Ibs.  per  sq.  in.  for  n  =  12,  15,  and  18  respec- 
tively. The  diagram  (a)  of  Plate  V  applies  also  to  this  one. 

For  working  stresses  q  is  taken  as  J  and  n  =  15.  In  experi- 
mental beam  tests  these  factors  should  be  previously  known. 

PROBLEMS 

28.  A  reinforced  concrete  beam  has  b  =  10  inches,  effective  d  =  14 
inches,   and   four  steel   rods   f-in.   diameter  in   the   tension  side,  two 
inches  from  the  bottom.     If  the  ultimate  compressive  strength  of  the 
concrete  be  2000,  and  the  elastic  limit  of  the  steel  be  40,000  Ibs.  per  sq. 
in.,  find  the  ultimate  resisting  moment  of  the  beam. 

29.  If,  in  the  last  problem,  the  safe  working  stresses  in  the  concrete 
and  steel  be  limited  to  600  and  16,000  Ib.  per  sq.  in.,  compute  the  safe 
resisting  moment  of  the  beam. 

30.  A  beam  12  feet  long  of  1:2:4  concrete  is  to  be  designed  to  safely 
withstand  a  bending  moment  of  150,000  Ib.  in.     Consult  Chapter  III. 
Make  assumptions  as  to  C  and  St,  and  compute  6,  d,  and  p. 

31.  Solve  the  last  three  problems  by  means  of  the  "  straight  line" 
methods  where  such  solution  is  practicable. 

32.  Solve  the  above  problems  if  practicable  by  the  "  full  parabolic  " 
relation,  and  tabulate  the  results  of  the  three  methods  of  solution. 

33.  In  Plate  V  plat  other  curves  for  p  for  five  values  of  q. 

Diagonal  Tension  in  Beams.  When  a  horizontal  beam  is 
subjected  to  vertical  loads,  there  is  a  tendency  in  each  small 

particle  to  move,  both  vertically  and 
horizontally,  with  respect  to  the 
particle  adjacent  above,  below,  and 
'  at  the  ends.  Beams  have  been  built, 
for  experimental  purposes,  of  brick 
resting  on  a  strap  of  iron,  as  shown 
in  (a),  Fig.  25.  Under  a  load  the 
tendency  is  to  assume  a  position  like 
(6),  that  is,  for  bricks  to  move  verti- 
cally with  respect  to  the  ones  adja- 
cent. This  represents  the  common  conception  of  vertical  shear. 


B 


ANALYSIS  OF  STRESSES  69 

When  a  pile  of  boards  or  a  pack  of  cards  is  bent,  as  shown  in 
Fig.  26,  each  board  or  card  slips  upon  the  one  above  or  below 
it.  This  action  is  also  that  of  shear, 
and  is  noticeable  in  beams  of  lami- 
nated material.  Deep  floor  beams  of 


hard  pine  frequently  fail  by  splitting     T  a 

from    end    to   end   in   a   horizontal 

plane,  midway  between  the  top  and 

bottom.     The  effect  of  the  splitting 

is  to  increase  the  deflection  beyond 

safe  limits,  although  the  tension  or 

compression  of  the  outer  fibers  may 

not   be   excessive.     In   general,    the  pIG  26 

upper  and  lower  fibers  of  beams  are 

subjected  to  greater  stresses  than  occur  in  the  interior  of  the 

same,  and  only  when  the  material  is  such  that  it  fails  under 

comparatively  light  stresses,  in  directions  other  than  horizontal, 

is  the  shear  of  relatively  great  importance. 

This  condition  exists  in  reinforced  concrete  beams  where 
failure  by  diagonal  tension  occurs  at  some  distance  from  the 
outer  fibers,  while  the  latter  are  sustaining  materially  larger 
stresses.  The  reason  clearly  is  that  the  concrete  is  stronger  in 
compression  than  in  tension,  and  that  the  strength  of  the  lower 
or  tension  edge  is  supplemented  by  the  steel. 

Distribution  of  Stresses  in  a  Homogeneous  Beam.  That  there 
are  horizontal  and  vertical  forces  acting  as  couples  on  the  sides 
of  all  small  cubical  particles  in  a  beam,  is  a  well  known  prin- 
ciple of  mechanics.  This  principle  is  illustrated  in  Fig.  26,  and 
Fig.  27  shows  that,  for  equilibrium,  the  horizontal  and  vertical 
unit  shears  must  be  equal  to  each  other.  The 
vertical  shear  is  readily  computed  from  the  loads 
on  the  beam,  and  the  average  unit  shear  is  this 
value  divided  by  the  area  of  the  cross-section. 
This  average  unit  shear  is,  however,  not  the 
maximum ;  or,  in  other  words,  the  unit  shear  varies 
from  the  outer  fiber  to  the  neutral  axis. 

Let  the  beam  in  Fig.  28  be  cut  by  two  vertical  sections  distant 
dx  apart,  and  let  the  sum  of  the  stresses  above  any  horizontal 
plane,  as  mn,  be  H  and  H'.  In  general  H  and  Hf  will  be  un- 
equal, and  np  will  have  a  tendency  to  move  toward  om,  causing 


70 


REINFORCED  CONCRETE 


shearing  along  mn.  Let  z  be  the  distance  from  the  neutral 
axis  to  any  fiber,  as  mn  of  cross  area  da,  and  b  be  the  breadth  of 
the  beam  at  this  point.  Then,  from  the  ordinary  flexural 
formula,  the  unit  stress  at  o  is  S  =  Me/ 'I,  and  at  p  is  S  = 
M'c/I,  where  c  is  the  distance  from  the  neutral  axis  to  the  ex- 
treme fiber,  and  the  stresses  on  a  fiber  of  da  section,  distant  z 
from  the  neutral  axis,  will  be  M/I-da-z  and  M'l-da-z.  The 
summation  of  these  stresses  for  values  of  da  between  z  and  c 

will  give  H  and  H'  respec- 
tively. The  difference,  H'  — 
H,  is  the  shear  on  the  surface 
mn-b  or  bdx,  and  M'  —  M  = 
dM.  If  v  be  the  unit  shearing 
pIG  28.  stress  on  mn-b  and  V  the  total 

vertical  shear  at  the  section, 

V  =  dM  -r-  dx  =  (Mf  —  M)  dx,  as  found  above,  and  v  =  (Hr 
—  H)  -r-  bdx.  Using  these  values,  the  expression  for  v  becomes 


1         1  1 

I 

•**            1              1          -H 

i 

T 

in      n 

1. 

z 

d  - 

i 

-IT-  I    daz 


or 


v  = 


Vm 
Ib 


(56) 


(57) 


where  m  is  the  statical  moment  with  reference  to  the  neutral 
axis,  of  the  part  of  the  cross-section  between  the  outer  fibers 
and  those  distance  z  from  the  neutral  axis.  Formula  (56) 
applies  to  any  form  of  cross-section  in  which  b  is  the  width 
at  ?.  from  the  neutral  axis.  In  case  the  section  is  rectangular 
the  value  of  the  unit  shear,  at  the  neutral  axis,  becomes 

,  V 
V==*bd 

or  |  of  the  average  shear.  It  is  seen  that  v  is  zero  when  V  is 
zero,  or  at  a  dangerous  section,  that  it  increases  as  the  section 
is  nearer  the  end  of  the  beam,  and  vertically, 
the  maximum  is  at  the  neutral  axis  since  m 
is  there  maximum. 

If  the  unit  stresses  for  a  vertical  rectangular 
section  be  plotted,  the  curve  so  formed  will 
be  a  parabola,  as  shown  in  Fig.  29,  where 
the  maximum  abscissa  is  on  the  neutral  axis. 


FIG.  29. 


ANALYSIS  OF  STRESSES 


71 


Shear  and  Axial  Stress.  If  the  horizontal  and  vertical  shear, 
v,  and  the  axial  stress,  C  or  St,  be  resolved  into  components 
parallel  and  perpendicular  to  the  diagonal  of  the  elementary 
prism,  the  latter  component  is  the  diagonal  tension.  When  this 
is  maximum  the  value  is  t  = 


S  +  Vv2  H-  i  £2,  and  it  makes 
an  angle,  0,  with  the  neutral  axis,  such  that  cot  20  =  S  -f-  2  v. 
The  directions  of  these  maximum  stresses  are  shown  in  Fig.  30. 
The  curves  give  the  directions  of  the  stresses  only  and  are  not 
lines  of  equal  stresses.  If  the  beam  were  to  fail  by  diagonal 
tension,  fracture  would  take  place  in  sections  perpendicular  to 
these  curves,  and  hence  reinforcing  rods  should  follow  approx- 
imately the  lines  in  Fig.  30.  Since  the  bending  moment  and 
the  tensile  stresses  decrease  from  the  middle  toward  the  ends 
of  a  beam,  it  follows  that  fewer  horizontal  reinforcing  rods  are 
required  near  the  extremities  than  where  the  moment  is  greater. 
The  rods  may  be  of  different  lengths,  stopping  short  of  the 
ends,  and  still  give  sufficient  horizontal  support,  but  it  is, 


FIG.  30. 

rather,  the  custom  to  bend  them  upward  at  proper  intervals 
to  reinforce  for  diagonal  tension.  In  addition,  vertical  rods, 
somewhat  shorter  than  the  depth  of  the  beam,  are  often  inserted 
for  the  purpose.  It  is  readily  seen  that  this  is  not  an  ideal 
arrangement  since  they  do  not  follow  the  direction  of  the  stresses, 
but  as  it  is  easier  to  so  place  them  than  to  fasten  them  with 
the  proper  slant  a  greater  number  may  be  used  at  the  same 
expense. 

An  inspection  of  the  formulas  for  t  and  6  show  that  where  S 
is  zero,  t  =  v,  and  the  direction  of  the  stress  is  45°  with  the 
neutral  axis.  This  is  the  condition  at  the  end  of  the  beam,  or 
at  a  point  of  inflection,  and  along  the  neutral  axis  where  v  is 
the  only  stress.  When  v  is  zero  0  is  zero  and  S  is  the  only  stress 
acting,  as  is  the  case  along  the  upper  and  lower  edges,  and  at 
any  point  of  maximum  moment. 

If  v,  S,  and  t  be  computed  and  plotted,  as  in  Fig.  31,  it  will 


72 


REINFORCED  CONCRETE 


675 


1200 


FIG.  31. 


be  seen  that  the  influence  of  shear  is  not  very  marked  in  beams 
of  normal  dimensions  and  loading,  and  may  be  neglected.  This 
applies,  however,  only  to  beams  of  homogeneous  material,  and 
the  effect  of  diagonal  tension  is  of  great  importance  in  reinforced 
concrete  construction.  To  illustrate  the  effect  of  shear  upon 

the  stresses  in  various 

— 50in.=  7/2—  ^          parts    of    the    vertical 

longitudinal  section, 
let  a  beam  be  assumed 
of  breadth  1  inch, 
depth  20  inches,  and 
length  100  inches,  with 
a  uniform  load  of  100 
pounds  per  linear  inch. 
In  Fig.  31  are  shown 
values  of  v,  St,  and  t,  in 

the  lower  half  of  the  beam,  at  the  lower  fibers,  at  the  neutral 
axis,  and  midway  between,  and  on  vertical  sections  10  inches 
apart.  At  the  bottom  v  is  zero;  at  the  neutral  axis  St  is  zero. 
On  the  mid-section  between,  t  is  given  above  the  line,  with  v 
and  St  in  order  below.  Some  lines  have  been  drawn  joining 
points  of  equal  stress,  but  it  will  be  noted  that  the  maximum 
stresses  are  in  the  lower  fibers. 

PROBLEMS 

34.  At  what  places  in  a  beam  are  the  stresses  zero? 

35.  If,  in  addition  to  the  uniform  load,  the  above  beam  be  loaded 
with  1000  pounds  at  20  inches  from  each  end,  what  changes  will  result 
in  Fig.  31?     Compute  and  plot  values  of  t. 

36.  What  would  be  the  nature  of  the  stress  (a)  in  a  horizontal  rod 
through  the  load  in  Fig.  25?     (6)  in  a  vertical  rod  in  Fig.  26? 

Distribution  of  Stresses  in  Reinforced  Beams.  While  the 
above  discussion  and  formulas  refer  to  homogeneous  beams, 
they  may  be  applied  to  reinforced  beams  if,  for  the  steel,  there 
be  substituted  enough  concrete  to  take  its  place  in  resisting 
deformation.  As  the  deformation  is  proportional  to  E,  it  fol- 
lows that  n  times  the  area  of  the  steel  displaced  must  be  added. 
A  conception  of  this  substitution  is  shown  in  Fig.  43,  where  the 
holes  left  by  the  steel  are  filled  by  the  concrete,  and  n  —  1  times 
as  much  is  added  at  the  same  distance  from  the  neutral  axis, 


ANALYSIS  OF  STRESSES 


73 


as  was  the  steel.  Then  the  moment  of  inertia  may  be  computed 
and  used  in  (57)  for  any  cross-section.  When  the  stresses  are 
increased  so  that  the  concrete  in  tension  is  broken,  different 
conditions  arise,  which  will  be  considered  below. 

In  Fig.  27  h  is  the  difference,  H'  -  H  in  Fig.  28,  if  the  former 
figure    be    applied    to    the    whole    cross-section,   and  v  be    the 
vertical    shear,  then   the   arm   of  h  is  the   distance   from   the 
center  of  gravity  of  the  steel  to  the  point  of 
application  of  the  compressive  stresses  in  the 
concrete,  or  j  in  (12).     Then  Vdx  =  hj,  h  being 
infinitesimal.     Also,  since  Hr  —  H  is  the  total 
shear  on  a  section  above  the  steel,  as  mn  in 
Fig.   28,  h  =  vbdx.      Making    an   equation   of 
the  two  values  of  h/dx,  there  results,  at  the 
neutral  axis  T/ 


v  = 


^ 

¥ 


(58) 


FIG.  32. 


In  general,  v  is  the  vertical  shear  divided  by  the  distance 
between  the  centers  of  compression  and  of  tension,  whatever 
the  shape  of  the  cross-section  or  distribution  of  reinforcement 
may  be. 

Since  H'  —  H  is  uniform  from  the  steel  to  the  neutral  axis 
when  there  is  no  tension  in  the  concrete,  the  value  of  v  remains 

unchanged  below  the  neutral  axis,  and 
the  diagram  for  v  over  the  entire  depth 
is  like  Fig.  33.  The  upper  part  of  the 
figure  is  determined  as  though  the  beam 
were  not  reinforced.  The  value  of  j 
may  be  taken  approximately  as  0.875  d 
for  rectangular  beams,  and  as  d  —  J  t  in 
T-beams.  In  T-beams  b  is  the  breadth 
of  the  web. 

In  computing  the  horizontal  stresses  in  beams  it  is  customary 
to  neglect  the  tensile  strength  of  the  concrete,  as  this  strength 
is  destroyed,  near  the  points  of  maximum  moment,  under  com- 
paratively light  loads.  This  is  proper  practice,  but  an  inspec- 
tion of  beam  failure,  such  as  shown  in  Fig.  34,  makes  it  plain 
that  the  steel  and  concrete,  acting  together,  fail  as  a  beam  along 
lines  similar  to  those  of  maximum  stress  in  homogeneous  beams. 
It  is  the  practice  to  counteract  this  tendency  to  weakness  by 


74 


REINFORCED  CONCRETE 


inserting  reinforcement  in  suitable  fashion  to  prevent  the  open- 
ing of  such  cracks.  Typical  beam  cracks  and  two  common 
styles  of  reinforcement  to  resist  diagonal  tension  are  shown  in 
Fig.  35,  (a)  and  (6).  In  (a)  a  rod  that  is  unnecessary,  to  the 
left  of  b  as  horizontal  reinforcement,  is  bent  up  at  an  angle  of 
about  45°  to  cross  the  possible  cracks  as  shown.  The  same 
result  is  attempted  in  (b)  where  rods,  called  stirrups,  are  placed 
vertically  and  looped  around  the  horizontal  rods.  If  the  beam 
were  of  wood,  a  crack  such  as  shown  might  be  closed  by  nails 


FIG.  34. 

driven  where  the  stirrups  or  the  diagonal  rods  are  indicated, 
both  the  nails  and  the  reinforcement  being  stressed  longitudinally 
rather  than  across  their  sections. 

Usually  with  diagonal  or  vertical  rods  the  assumption  is  that 
a  rod  as  //'  takes  a  part,  and  that  the  concrete  takes  the  re- 
mainder, of  the  average  vertical  shear  over  the  space  s,  repre- 
sented by  the  component  of  this  shear  in  the  direction  of  the 
rod.  The  usual  assumption  is  that  the  steel  and  the  concrete 
act  together,  rather  than  separately,  in  resisting  shear.  Prac- 
tically, a  definite  part  of  the  shear  is  assigned  to  a  part  of  the 
shearing  strength  of  the  concrete  and  the  rod  is  designed  to 


ANALYSIS  OF  STRESSES  75 

carry  the  remaining  vertical  shear  normal  to  the  surface  bs, 
where  b  is  the  breadth  of  the  beam.  The  tension  in  the  rod  is 
distributed  by  the  remaining  shearing  strength  of  the  concrete 
through  the  distance  \  s  toward  the  adjacent  rods.  It  is  well 
proved  by  experiments  on  beams  with  web  reinforcement  that, 
for  even  greater  than  safe  working  stresses,  the  concrete  is 
capable  of  taking  a  material  part  of  the  shearing  and  diagonal 
stresses;  and  that  not  more  than  two  thirds  of  the  vertical 
shear,  due  to  loads,  need  be  taken  care  of  by  the  web  reinforce- 
ment, the  other  third  being  well  within  the  safe-working  shearing 
strength  of  the  concrete. 

Working    Shearing    Stresses.     The    recommended    limits    of 
shearing  stresses  to  be  allowed  are  stated  on  page  150,  and  are 
based  upon  observations 
on   beams  composed   of 
concrete  capable  of  hold- 
ing 2000  Ib.  per  sq.  in.  in  (a) 
compression  at  the  end  of 
28  days.     The  first  rule 
(a)  applies  to  rather  thin 
slabs  and  to  small  beams 
of  rectangular  cross-sec-                              VpiG  35 
tion.1     Such  beams  may 

frequently  be  designed,  with  horizontal  steel  only,  without 
exceeding  the  limit  of  40  pounds  per  square  inch.  Large  beams 
and  girders  are  usually  of  the  T-section,  and  the  cross  area  is 
much  reduced.  As  a  result  the  limit  named  in  (a)  is  nearly 
always  exceeded  and  web  reinforcement  is  introduced.  Where 
the  horizontal  reinforcement  is  made  up  of  a  number  of  rods 
such  that  they  may  be  bent  up  in  pairs  at  frequent  intervals, 
the  web  strength  of  the  beam  is  obviously  increased  and  the 
unit  shearing  strength  of  the  cross-section  of  the  beam  is  greater 
than  before.  When  the  number  of  such  bars  to  be  bent  up  is 
such  that  every  probable  crack  will  be  crossed  by  at  least  one 
of  them  below  the  neutral  axis,  the  maximum  shearing  stress 
on  the  area  of  the  cross-section  may  be  increased  to  60  Ib.  per 
sq.  in.  according  to  rule  (b).  If  the  bent-up  rods  or  vertical 
stirrups  be  spaced  f  of  the  depth  apart,  the  maximum  shearing 
stress  may  be  taken  as  120  Ib.  per  sq.  in. 

1  See  page  79  for  discussion  of  this  question. 


V 

,''\e    f 


76  REINFORCED  CONCRETE 

Computation  of  Stresses  in  Web  Reinforcement.  While,  in 
general,  such  stresses  are  not  capable  of  exact  computation, 
approximate  values  may  be  found  that  are  on  the  side  of  safety. 
In  Fig.  36  (a),  the  average  vertical  shear,  due  to  external  loads 
on  the  beam  over  the  space  i  s  on  each  side  of  a  vertical  stirrup 
is  carried  in  part  by  the  concrete,  and  in  part  by  the  steel.  If 
V  be  the  shear  not  apportioned  to  the  concrete,  the  maximum 
unit  shear  will  be  v  =  V  -J-  bj,  or  about  V  -*-  0.875  bd.  This 

is  the  vertical  component 
of  the  unit  diagonal  ten- 
sion and  causes  tension  in 
the  vertical  reinforcement. 
If  the  section  area  of  all 
the  stirrups,  acting  in  a 

right  section,  be  a,  and  the  unit  tensile  stress  be  St,  the  whole 
load  on  the  rods  will  be  aSt,  and 

aSt  =  vbs  =  -^  (59) 

For  usual  values  of  St  and  j,  in  rectangular  beams,  this  becomes 


It  is  clear  that  vertical  reinforcement  is  ineffective  to  prevent 
vertical  cracks  such  as  appear  under  very  small  loads,  but  that 
they  become  more  efficient  as  the  cracks  incline  more  toward 
the  middle.  In  case  the  beam  is  of  the  T-section,  formula  (59) 
is  used. 

When  the  rods  are  placed  at  an  angle,  a,  with  the  horizontal, 
the  distance  between  the  rods  is,  for  the  same  horizontal  spac- 
ing, s  sin  a,  and  the  tension  on  each  rod  becomes 

aSt  =  -r^  sin  a  (62) 

The  rods  are  usually  bent  up  at  an  angle  of  45°,  especially 
where  several  pairs  are  inclined;  then,  for  usual  values  of  S 
and  j,  in  rectangular  beams, 

a  =  T™  (63> 


ANALYSIS  OF  STRESSES  77 

19,800  ad 
and  s  =  --*-jj-  (64) 

To  investigate  a  design  the  solution  for  St  is  made  from  (59). 

The  vertical  shear  in  these  formulas  is  usually  considered 
as  two-thirds  that  occasioned  by  the  loads  on  the  beam,  the 
other  third  being  taken  by  the  concrete. 

Spacing  Web  Reinforcements.  The  spacing  of  the  stirrups  or 
bent-up  rods  may  be  limited  otherwise.  As  stated  above,  a 
diagonal  tension  crack  should  always  be  crossed  by  a  piece  of 
web  reinforcement  below  the  neutral  axis.  This  obviously 
requires,  with  vertical  stirrups  and  «  assumed  as  45°,  that  the 
horizontal  spacing  shall  be  less  than  0.6d  for  rectangular 
cross-sections,  and  less  than  d  —  t  for  T-beams.  As  most  of 
the  heavy  beams  are  of  the  latter  form,  it  is  common  practice 
to  make  the  maximum  spacing  not  more  than  f  d,  as  is  recom- 
mended on  page  151. 

If  values  of  s  in  (59)  and  (62)  give  spacing  less  than  f  d,  the 
smaller  distance  must  be  used.  This  is  not  at  all  unusual  in 
deep  beams.  For  example,1  slabs  30  inches  deep  carrying  heavy 
railway  traffic  over  23-foot  spans  in  Chicago,  have  the  U-shaped 
stirrups  18  inches  apart. 

Points  for  Bending-up  Bars.  Since  the  bending  moments  in 
a  beam  decrease  from  the  maximum  near  the  middle  to  zero 
at,  or  near,  the  ends,  it  follows 
that  the  same  amount  of  horizontal 
reinforcement  is  not  required 
throughout  the  whole  length  of  the 
beam. 

If  M  be  the  maximum  bending 
moment  of  a  beam  fixed  at  the  ends, 
MI  be  the  moment  at  any  other 

point  as  6,  wx  the  uniform  load  between  a  and  6,  and  wl/2 
the  shear  at  the  left  support 

Mi  =  M  -  J  wx2 
from  which  the  bending  moment  at  any  point  may  be  computed. 

In  reinforced  beams  extending  over  several  spans  there  is 
more  or  less  continuity  over  and  fixedness  at  the  supports.  As 
it  is  unknown  at  the  time  of  making  the  design  what  the  exact 

1  For  details  of  tests  of  three  of  these  beams  see  Bulletin  No.  28,  Univer- 
sity of  Illinois  Engineering  Experiment  Station,  1908. 


78 


REINFORCED  CONCRETE 


distribution  of  the  loads  will  be,  it  is  usual  to  assume  the  maxi- 
mum positive  bending  moment  to  be  yV  wl2  at  the  middle  of  all 
beams  except  the  end  spans  where  it  is  10  wl2.  The  maximum 
negative  bending  moments  are  TV  wl2  at  all  supports  except 
those  at  the  ends  where  the  moment  is  yV  wl2.  If  the  diagrams 
of  bending  moments  be  assumed  to  be  parabolas  with  origins 
at  the  mid-sections  of  the  beams,  the  height  of  the  diagrams 
being  the  sums  of  the  positive  and  negative  moments,  the  curves 

T 


Distance  from  Middle  of  Beam  to  Bend 


FIG.  38. 

in  Fig.  38  may  be  plotted  and  are  to  be  used  in  finding  the 
points  at  which  any  number  of  rods  may  be  dispensed  with, 
in  the  tension  side,  and  bent  up  for  web  reinforcement.  For 
example,  let  the  reinforcement  consist  of  2  one-inch,  2  seven- 
eighth-inch,  and  2  three-quarter-inch  rods,  arranged  symmetri- 
cally with  the  larger  ones  outside.  The  whole  cross  area  of 
the  steel  is,  then,  from  page  156,  3.657  square  inches.  The 
pair  in  the  middle  of  the  breadth,  to  be  bent  up  nearest  the 
middle  of  the  beam,  have  an  area  of  0.8836  square  inches 
or  0.24  of  the  whole.  Hence,  where  the  bending  moment 
is  but  0.76  of  the  maximum,  the  other  four  rods  are  suffi- 
cient to  provide  for  it.  In  Fig.  38,  for  M  =  T*T  wl2,  it  is  seen 
that  0.24  of  the  area  of  the  rods  may  be  bent  up  at  a 
point  0.17 1  from  the  middle.  The  other  two  rods  to  be 
bent  up  have  an  area  of  1.203  square  inches,  or  there  will 
be  left  1.57  square  inches  or  0.43  of  the  whole.  In  Fig.  38 
it  is  seen  that  0.57  of  the  steel  may  be  bent  up  at  0.27  I  from 


ANALYSIS  OF  STRESSES  79 

the  middle.  The  two  outside  rods  run  the  whole  length  of  the 
beam.  Usually  the  rods  are  continued  a  little  farther  than 
the  minimum  distance,  bent  up  to  a  point  near  the  top  of  the 
beam  and  again  bent  to  a  horizontal  position,  where  they  become 
tensile  reinforcement  to  resist  the  negative  moment  near  the 
supports.  This  method  is  not  quite  accurate  since  the  resisting 
moment  of  a  beam  does  not  vary  directly  as  the  amount  of 
reinforcement.  However,  the  above  process  makes  the  stress 
in  the  rods  nearly  constant  but  decreasing  from  the  middle, 
while  the  stress  in  the  concrete  is  also  decreased  from  that  at 
the  middle,  but  not  directly  with  the  bending  moment. 

As  determined  by  (2),  u  =  V  -f-  moj,  a  certain  number  of 
rods  must  always  run  to  the  ends  of  a  beam  whether  they  are 
needed  to  resist  flexure  or  not. 

Beams  not  Requiring  Web  Reinforcement.  Most  reinforced 
concrete  beams  of  rectangular  cross-section  are  in  the  form  of 
comparatively  thin  and  wide  slabs.  These  slabs  frequently 
form  the  flanges  of  T-beams,  and,  as  such,  are  entirely  in  com- 
pression, but  between  the  T-beams  they  act  as  simple  beams. 
Unless  the  slabs  are  of  unusual  depth  the  concrete  may  be  able 
to  take  all  the  diagonal  stresses,  and  stirrups  may  be  unnecessary. 
It  may  be  readily  ascertained  whether  or  not  this  is  the  case 
when  the  depth  and  span  are  known.  From  (16),  page  40, 
M  =  NiCbd2,  and,  for  uniform  loading, 

cwl2  = 

Also  V  =  i  wl 

wl 


or 


2bj 


,.   .  I     N,Cd 

combining,  g=—  (65) 

in  which  c  is  the  coefficient  depending  upon  the  way  in  which 
the  ends  of  the  beams  are  fixed.  For  a  simple  beam  c  is  J.  If 
j  be  0.875  d,  v  be  40,  St  be  16,000,  and  C  be  650  pounds  per 
square  inch,  Ni  is  0.166  from  Plate  I,  and 

^  =  -y-  =  12.3  for  simple  beams. 

If  c  be  TV,  l/d  =  15.4,  and  for  c  =  TV,  l/d  =  18.5. 

For   concrete   of   strength   such   as   assumed,    stirrups   must 


80  REINFORCED  CONCRETE 

therefore  be  provided  if  the  length  be  not  more  than  12.3  d 
for  simple  beams,  and  15.4  d  or  18.5  d  for  beams  having  ends 
partially  or  entirely  fixed. 

Bond  Stresses  in  Web  Reinforcement-stirrups.  It  is  evident 
that  for  economy  of  design  the  bond  between  the  web  reinforce- 
ment and  the  concrete  should  be  equal  to  the  strength  of  the 
rods.  As  was  shown  on  page  25,  the  length  of  grip  of  the  rod 

should  be  ,  e 

x  =  di&t  -T-  4  u. 

Substituting  the  value  of  St  from  (87)  this  becomes 

Vt 

x  =  -^-  (66) 


x  = 


in  which  m  is  the  number  of  stirrups  in  the  area  6s.  It  will 
be  noted  that  x  varies  inversely  as  dim.  This  expression  applies 
to  square  bars  if  4  be  substituted  for  TT.  For  ordinary  working 
values,  s  may  be  taken  as  f  d,  j  as  .875  d,  and  u  from  75  to  80 
pounds  per  square  inch  of  contact  between  steel  and  concrete, 
then  for  round  rods 

0.273  V  0.215  V  , 

j—    -  or  x  =  —  i—    -  for  square  bars  (67) 

d\mu  a\mu 

0.0035  V  0.0028  V  , 

or  x  —  —  j  —    -  or  x  =  —  j  —    -  for  square  bars          (68) 

d^m  d^m 

In  this  expression  V  is  the  vertical  shear  not  taken  by  the 
concrete.     If  V  be  the  shear  due  to  loads  the  value  of  x  is 

.0035(7  -  35  6d), 
x  =  -        -^  -        -  for  rods  (69) 

0.0028  V  , 
~dw  ~   °F  bai>S  (70) 

When  the  safe  shearing  strength  of  the  concrete  is  40  Ib.  per 
sq.  in.,  and  that  of  the  whole  beam  120  Ib.  per  sq.  in.,  or  when 
V  is  I  7, 

0.0023  V  0.0018  Vt     , 

x  =  -—j  —    -  or  x  =  —  -=  --  for  bars. 
djin  <Ji»i 

When  the  stirrups  are  vertical  the  length  is  limited  by  the  depth 
of  the  beam.  Since  the  tension  side  of  the  beam  is  apt  to  con- 
tain many  fine  vertical  cracks  it  is  not  reasonable  to  expect 
full  bond  strength  throughout  the  entire  depth.  Experiments 


ANALYSIS  OF  STRESSES  81 

show  that  it  is  safe  to  assume  the  grip  of  a  rod  to  be  T%  of  its 
length  in  computing  bond  strength. 

In  designing,  all  the  factors  in  (66)  are  generally  known  or 
assumed  except  di  and  m,  the  product  of  which  may  be  found 
from  the  same  equation: 

0.0035  V  0.0028  V 

mdi  =  -  — )  or  for  squares  -    — ^ — 

x  d 

0.006  Vf  0.0046  V 

=  -  —3 >  or  for  squares  -    —3 

d  d 

0.006(7-  35  bd)  , 

--  for  round  rods 


d 

0.0046(7  -  35 
d 


for  square  bars 


0.004  V        ,  0.003  V 

and  mdi  — -= >  or  for  squares  — -j—  (71) 

when  the  concrete  takes  J  and  the  stirrups  take  f  of  the  vertical 
shear.  If  rods  of  several  sizes  be  used  mdi  is  the  sum  of  the 
diameters. 

For  example,  let  it  be  required  to  reinforce  a  beam  against 
diagonal  tension  when  6,  d,  and  I  are  10  inches,  20  inches,  and 
10  feet  respectively,  and  the  uniform  load  is  2000  Ib.  per  linear 
foot.  The  maximum  shear  is  2000  X  5  =  10,000  pounds.  The 
concrete  will  take  35  X  10  X  20  =  7000  pounds,  hence,  V  is 
3000  pounds.  Then  mdi  =  .006  X  3000  •*•  20  =  0.9  inch,  or  4 
quarter-inch  rods.  At  the  ends  the  spacing  will  be  not  greater 
than  s  =  14,000  X  0.25  X  20  -f-  3000  =  23  inches,  but  a  is  other- 
wise limited  to  f  of  20  =  15  inches.  If  two  half-inch  rods  be 
used,  s  is  computed  to  be  47  inches,  which  arrangement  is  less 
economical.  By  using  even  smaller  rods  the  computed  value 
of  s  could  be  made  to  agree  with  the  specified  limit,  f  d;  such 
sizes  are,  however,  unusual  for  this  purpose.  The  next  stirrups 
are  15  inches  nearer  the  middle,  where  they  carry  but  500  pounds 
of  shear  and  but  two  rods  need  be  used. 

Bond  Stresses  in  Bent-up  Rods.  If  the  above  expressions 
for  x  and  dim  be  multiplied  by  sin  a,  they  will  apply  to  rods 
inclined  at  that  angle  with  the  horizontal.  Usually  the  only 
case  of  inclined  web  reinforcement  is  that  of  the  bent-up  bars 


82 


REINFORCED  CONCRETE 


of  the  horizontal  reinforcement.  In  order  that  they  may  be 
bent  up  at  frequent  intervals,  the  rods  should  be  small  and 
numerous  rather  than  large  and  few,  which  arrangement  also 
provides  better  bond. 

Transverse  Spacing  of  Reinforcement.  In  order  that  there 
shall  be  beam  action  it  is  necessary,  not  only  that  there  be  per- 
fect bond  between  the  steel  and  the  adjacent  concrete,  but  also 
that  the  steel  and  adjacent  concrete  shall  not  shear  away  from 
the  remainder  of  the  concrete.  This  condition  requires  that  the 
bond  between  the  steel  and  the  concrete  be  equal  to  the  shearing 
strength  of  the  concrete  between  the  rods.  Only  the  bond  on 
the  surface  of  the  rod  above  the  plane  of  least  concrete  is  effective 
in  causing  shear  in  that  plane,  the  bond  on  the  remainder  of 
the  rod  being  exerted  upon  the  concrete  below.  Thus  in  (a), 
Fig.  39,  i  *diu  =  sSs,  in  (6),  dm  =  sSs,  and  in  (c),  2  diu  =  sSs. 
As  the  shearing  strength  of  concrete  may  be  safely  taken  at 
more  than  three  halves  that  of  bond,  the  values  of  s  in  the 

above  expres- 
sions become 
1.05  di,  0.7  rf,, 
and  1.33  di. 
The  last  is 
an  unusual  ar- 
rangement, and  if  the  diagonal  instead  of  the  side  be  taken  as 
the  thickness  of  the  bar,  the  clear  width  of  the  concrete  will 
be  0.94  times  the  diagonal. 

The  reinforcement  is  usually  put  in  place  within  the  forms 
before  the  concrete  is  poured,  and  it  is  essential  that  enough 
space  be  left  between  the  rods  to  allow  the  concrete  to  entirely 
surround  the  steel  and  to  fill  the  forms.  Between  the  steel  and 
the  outside  of  the  beam  the  concrete  acts  as  a  protection  to  the 
reinforcement  against  fire  and  dampness.  For  these  purposes 
it  is  not  possible  to  prove  very  definitely  the  thickness  needed, 
but  the  recommendations  of  the  Joint  Committee  represent 
usual  practice.  See  page  144. 

Working  Rule  for  Transverse  Spacing  of  Reinforcement.  The 
recommended  practice  is  given  on  page  150,  and,  in  designing 
beams,  the  width  must  accordingly  be,  at  least, 


6  = 


2.5n)d,, 


ANALYSIS  OF  STRESSES  83 

Where  n  is  the  number  and  di  the  thickness  of  the  rods.  In 
addition  there  must  be  a  width  of  at  least 

b  =  4  +  2.5  (n  —  1)  di  for  girders, 
or  b  =  3  +  2.5  (n  —  1)  di  for  beams. 

For  ultimate  strength  of  the  beam,  a  greater  width  may  be  of 
advantage  since  adhesion  to  the  horizontal  rods  is  partly  de- 
stroyed and  the  strength  is  largely  dependent  upon  the  anchor- 
age at  the  ends.  This  anchorage  may  be  more  effectual  as  the 
width  of  the  beam  is  increased. 

The  Inclination  of  Bent-up  Bars.  At  a  point  in  a  beam  where 
the  bending  moment  has  decreased  from  the  maximum  suffi- 
ciently to  allow  a  part  of  the  horizontal  rods  to  be  dispensed 
with,  they  may  be  (a)  discontinued,  (6)  turned  over  to  form  a 
hook,  or  (c)  continued  diagonally  upward  toward  the  top  of 
the  beam.  The  first  alternative  is  not  desirable  as  the  bond 
in  the  concrete  near  the  middle  of  the  beam  is  not  as  good  as 
would  be  the  case  near  the  end  where  the  moments  are  smaller. 
Also,  when  a  part  of  the  bars  are  discontinued  the  stress  in 
those  remaining  becomes  immediately  increased,  which  fact 
impairs  the  bond  between  the  steel  and  the  concrete.  This 
latter  objection  applies  also  to  the  anchoring  of  the  rods  at 
such  points  by  hooks.  Usually  the  rods  are  bent  up  to  form 
diagonal  reinforcement.  This  arrangement  is  desirable  for 
several*  reasons:  the  horizontal  components  of  the  upturned 
bars  act  with  those  unbent  in  taking  the  tension  due  to  bending, 
and  so  the  resisting  moment  is  decreased  gradually,  as  it  should 
be.  Again,  such  a  rod,  being  rigidly  attached  to  the  horizontal 
part  and  extending  into  concrete  in  compression,  is  in  the  best 
possible  position  to  offer  resistance  to  diagonal  tension,  and  is 
put  in  place  with  less  extra  expense  than  is  a  detached  diagonal 
member  and  an  anchorage  for  the  horizontal  rod. 

The  horizontal  stress  in  a  rod  is,  at  the  bend,  transferred  to 
the  concrete  as  compression,  and  if  the  bend  be  too  abrupt, 
the  unit  stress  may  become  excessive. 
In  Fig.  40  r  is  the  radius  of  the  curve 
around  which  the  rod  is  bent,  and  the 
compression  stresses  may  be  represented 
by  some  such  figure  as  is  shown  in  the  JTIG.  40. 


84  REINFORCED  CONCRETE 

shaded  part.  If  the  length  of  the  curve  over  which  the  rod 
exerts  compression  be  assumed  to  be  r  and  the  unit  compres- 
sion be  C, 

Crd  =  7rd2St  +  4 

r  =  0.7854  dSt  +  C  (72) 

The  rod  is  imbedded  in  the  concrete  and  the  compression  is 
exerted  over  but  a  small  part  of  the  whole  cross-section.  Also 
the  concrete  is  normally  in  tension,  and,  for  these  reasons,  a 
relatively  high  stress,  as  1000  Ib.  sq.  in.,  may  be  assumed  for 
the  value  of  C.  If  the  working  strength  of  the  steel  be  taken 
as  16,000  pounds  per  square  inch,  the  stress  at  the  point  where 
the  bend  occurs  will  seldom  be  more  than  three-fourths  of  this, 
or  12,000  Ib.  sq.  in.  In  any  given  case  the  stress  may  be  com- 
puted. Substituting  these  values  of  St  and  C  in  the  above 
expression,  r  becomes 

r  =  .7854  X  12,000  d  -s-  1000  =  9.4  d. 

If  the  arc  in  compression  be  taken  as  45°,  r  becomes  12  d.  As 
cracks  following  the  line  of  the  rod  are  seldom  seen  in  tests 
of  beams  having  even  sharper  bends  in  the  reinforcement,  it 
may  be  assumed  that  curves  such  as  indicated  will  be  satis- 
factory. 

Waterproofing  and  Fireproofing  Qualities  of  Concrete.  As 
the  tension  side  of  a  beam  is  liable  to  contain  small  cracks  ex- 
tending past  the  reinforcement,  it  may  be  questioned  whether 
or  not  the  imbedded  steel  is  apt  to  become  rusty  and  subject 
to  corrosion  from  moisture  and  other  causes.  Most  of  the 
tests  and  observations  in  this  connection  have  been  made  on 
steel  imbedded  in  large  blocks  of  concrete,  and  kept  alternately 
under  water  and  in  air  for  various  periods  of  time.  These 
tests  usually  fail  to  show  any  marked  effect  of  the  water  upon 
the  steel  where  there  has  been  perfect  union  between  the  metal 
and  the  concrete.  In  general,  rusting  occurs  only  under  the 
combined  action  of  moisture  and  carbonic  acid.  The  preserving 
quality  of  concrete  is  then  dependent  upon  its  ability  to  exclude 
one  of  these  agents  from  contact  with  the  steel.  In  the  case 
of  large  blocks  of  concrete  under  compression  the  reinforcement 
may  be  perfectly  protected  by  the  fact  that  the  concrete  is 


ANALYSIS  OF  STRESSES  85 

water-tight.  In  the  tension  sides  of  beams,  however,  there  are 
always  some  cracks  which  may  become  filled  with  moist,  smoky 
air,  and  conditions  favorable  to  the  formation  of  rust  may  result 
if  the  metal  be  exposed.  As  concrete  contains  a  large  propor- 
tion of  lime,  it  readily  absorbs  and  neutralizes  the  carbonic 
acid,  and  effectually  protects  the  steel  if  it  completely  covers 
the  latter  with  even  a  thin  film.  In  order  that  this  film  may 
be  continuous  the  concrete  should  be  mixed  wet,  and  should 
contain  enough  fine  material  to  render  it  dense.  Bars  taken 
from  such  concrete  usually  have  a  coating  of  that  material  that 
is  removed  only  by  considerable  friction,  and  the  conclusion  is 
that  imbedded  steel  is  not  liable  to  rust  even  when  the  unit 
stress  in  it  approaches  the  elastic  limit.  Several  references  to 
tests  concerning  this  subject  are  noted  below.1 

The  Baltimore  and  San  Francisco  fires  and  numerous  smaller 
conflagrations  have  afforded  opportunities  for  the  study  of  the 
effect  of  great  heat  on  concrete  structures,  as  they  are  usually 
built.  In  addition,  some  tests  have  been  conducted  with  heat 
at  a  known  intensity,  and  other  causes  and  effects  more  accu- 
rately determined  than  can  be  the  case  in  a  burning  building. 
As  concrete  in  its  manufacture  has  passed  through  a  period  of 
intense  heat  it  suffers  but  little  from  the  further  application 
of  high  temperatures.  There  was,  however,  some  question  as 
to  the  effect  of  the  concrete  in  preserving  the  reinforcing  steel 
from  surrounding  fire. 

The  fireproofing  qualities  of  concrete  seem  to  be  due  to  the 
fact  that  it  is,  of  itself,  not  inflammable,  and  also  to  the  fact 
that  it  is  a  poor  conductor  of  heat.  The  mixture  in  setting  com- 
bines chemically  with  quite  a  large  proportion  of  water  which 
is  vaporized  by  heat  and  has  a  decidedly  cooling  effect ;  and  the 
material  so  changed  has  a  lower  thermal  conductivity  than 
before.  All  concrete  is  somewhat  porous  and,  the  entrained 
air  being  of  very  low  conductivity,  this  fact  adds  further  protec- 
tion to  the  metal.  In  this  respect  cinder  concrete  seems  some- 
what better  than  that  made  of  the  purer  quartz  sands. 

Both  the  experimental  tests  and  observations  of  results  of 
large  fires  have  been  fully  reported  and  only  the  summary  of 

1  Engineering  Record,  Vol.  LVII,  page  105;  Proc.  Am.  Soc.  Testing  Materi- 
als, Vols.  V,  VI,  and  VII;  Engineering  News,  August,  1904,  June  16,  1904, 
page  561. 


86  REINFORCED  CONCRETE 

conclusions  need  be  given  here.1  The  conclusions  to  be  drawn 
are  that  sharp  corners  of  columns  and  beams  are  more  suscep- 
tible to  attack  than  wide,  flat  surfaces,  such  as  slabs.  Ample 
protection  seems  to  be  afforded  by  2  to  2J  inches  and  f  to  1 
inch  of  concrete  in  the  two  cases  respectively.  Only  the  steel 
in  actual  contact  with  the  flame  seems  to  be  affected.  In  the 
first  reference  below  it  is  reported  that  1500°  Fahr.  for  two 
hours  at  the  surface  of  a  plate  resulted  in  only  500°  to  700°  Fahr. 
2  inches  under  the  surface,  while  at  3  inches  below  the  tempera- 
ture was  212°  Fahr.  Also,  the  temperatures  of  a  bar  protruding 
into  a  temperature  of  1700°  Fahr.  were  1000°  at  2  inches,  500° 
at  5  inches,  and  212°  at  8  inches  inside  the  surface. 

In  extreme  cases  reinforced  concrete  buildings  may  be  rendered 
valueless  by  fire,  but  the  destruction  will  be  accomplished  very 
slowly. 

Temperature  Stresses.  When  concrete  structures,  such  *as 
long  retaining  walls,  arches,  and  spandrel  walls,  and  continuous 
or  fixed  beams,  are  subjected  to  changes  of  temperature,  they 
will  tend  to  increase  or  decrease  in  length  according  to  a  known 
law.  If  the  resulting  stresses  be  sufficient,  the  concrete  will 
be  cracked  at  points  of  least  strength.  As  the  strength  of 
concrete  is  much  less  in  tension  than  in  compression,  the  results 
of  a  fall  in  temperature  are  more  serious,  usually,  than  in  case 
of  a  rise.  The  problem  of  reinforcing  for  temperature  stresses 
is  not  at  all  to  prevent  cracks,  but  only  to  cause  their  uniform 
distribution  along  the  entire  length  of  the  structure,  thus  keeping 
the  width  of  any  one  within  desired  limits.  The  coefficient  of 
contraction  of  dense  concrete  such  as  is  used  in  important  work, 
has  been  determined  2  with  considerable  accuracy,  and  may  be 
taken  as  .0000055.  The  range  of  temperature  in  concrete  is 
not  nearly  that  of  the  atmosphere,  and  a  drop  of  50°  Fahr.  is, 
perhaps,  all  that  need  be  allowed  for.  As  the  coefficient  of  ex- 
pansion of  steel,  0.0000065,  differs  but  slightly  from  that  of  con- 
crete, and  as  the  steel  is  somewhat  protected  by  the  covering 
of  concrete,  little  error  will  result  from  assuming  equal  change 
in  length  in  the  two  substances. 

1  See  Proc.  Am.  Soc.  Civil  Engineers,  Vols.  51,  55  and  59.     Cement,  May, 
1902,  page  95;  Eng.  News,  October,  1902,  page  334;  January,  1904,  page  30, 
March    24,    1904.     Insurance   Engineering,    1901,    page    483.     Engineering 
Record,  April  12,  1902,  page  341. 

2  Journal  Western  Soc.  Eng.,  Vol.  VI,  page  49. 


ANALYSIS  OF  STRESSES  87 

If  D  be  the  drop  in  temperature  from  normal  the  contraction 
is  0.0000055  Dl.  If  the  tensile  strength  of  concrete  be  300 
Ib,  sq.  in.,  and  E  be  2,000,000,  the  whole  elongation  is 

A  =  ~  =  0.0000055  Dl 


and  D  =  27°, 

so  cracks  will  occur  in  concrete  at  some  particularly  weak  spot 
on  the  surface  when  the  temperature  is  lowered  by  this  amount. 
With  increasing  cold  the  cracks  become  deeper  till  the  steel  is 
reached.  Here  there  can  be  no  opening  of  the  crack  until  the 
bond  between  the  concrete  and  the  steel  is  impaired.  As  the 
bond  at  the  crack  loosens,  the  concrete  slips  back,  the  bar 
stretches  and  its  cross-area  becomes  smaller.  This  action  im- 
pairs or  breaks  the  bond  farther  back,  and  the  bar  stretches 
over  the  space  y  on  each  side 
of  the  crack,  as  in  Fig.  41,  till 
equilibrium  is  established  be- 
tween the  stress  and  the  bond. 
Beyond  this  point  the  steel 
and  concrete  contract  equally, 
as  the  bond  is  perfect  over  the 
spaces  x,  which  extends  to  a 

point  y  distant  from  the  next  crack  or  to  the  immovable  wall  as 
shown.  If  there  be  but  one  crack  in  the  distance  I,  the  tensile 
strength  of  the  concrete  section  and  of  the  bond  are  equal.  As 
the  unit  tensile  and  bond  stresses  are  practically  equal,  or  may 
be  determined  for  any  particular  case,  an  equation  may  be  writ- 
ten: 2  x7rdiU/2  =  AS/p,  or  4  pAux  =  ASdi,  where  AS/p  is  the 
tensile  strength  of  the  concrete  and  u  is  somewhat  less  than  the 
ultimate  unit  bond  strength.  The  bond  stress  in  the  distance 
x  decreases  from  u  to  nearly  zero  at  the  wall,  and  the  average 
is  u/2  for  this  distance,  then  if  p  be  the  ratio  of  steel  to  concrete 
area  and  d\  the  diameter  of  the  rod  or  the  thickness  of  the  bar, 

--ft 

The  contraction  in  the  rod  in  the  space  x  is  the  same  as  the 
expansion  in  the  distance  y,  as  I  is  constant,  each  being  .0000055 


88 


REINFORCED  CONCRETE 


Dx,  or,  if  D  be  50°,  y  =  .000275  x.     If  the  elastic  limit  of  steel 
be  33,000  Ib.  sq.  in., 

33,000  y 
E 


and 
also 


so 


.000275  x  = 

x  =  4y, 


2.5 


T)    —  L 

P  ~  1.6  / 


(73) 


The  constants  will  vary  somewhat    for  different  concretes, 
but  it  will  be  noted  that  the  distance  between  cracks  is  inversely 


FIG.  42. 


proportional  to  the  amount  of  reinforcement.  Also,  that,  for 
economy,  the  reinforcement  should  be  made  up  of  small  rods, 
while  no  amount  of  reinforcing  will  prevent  cracks  or  diminish 
their  aggregate  width,  which  will  always  be  AD/,  when  A  is  the 
coefficient  of  expansion.  Fig.  42  is  plotted  from  (73)  for  several 
sizes  of  rods,  either  square  or  round. 

For  example,  let  the  railing  of  a  bridge  be  300  feet  long  and 
fixed  at  the  ends.  The  contraction  due  to  50°  fall  in  temper- 
ature will  be  .0000055  X  50  X  300  X  12  =  0.99  inches,  which 
amount  will  be  the  sum  of  the  widths  of  all  the  cracks.  If  there 
be  99  cracks,  each  0.01  inch  wide,  I  in  (73)  will  be  3  feet,  and 
from  Fig.  42  the  required  amount  of  reinforcement  is  found  to 
be  0.44  per  cent,  for  J-inch  rods,  or  0.88  per  cent,  for  half-inch 
rods.  If  0.5  per  cent,  of  half-inch  rods  be  used,  the  cracks  will 


ANALYSIS  OF  STRESSES  89 

be  5.2  feet  apart  and  the  width  of  each  will  be  0.99  X  5.2  -r- 
300  =  0.017  inches. 

If  the  concrete  contracts  in  hardening,  cracks  will  result  if 
the  tensile  strength  be  exceeded  by  the  resulting  stress.  As 
the  steel  is  otherwise  unstressed  it  will  resist  contraction  of  the 
concrete  and  be  in  compression  where  the  bond  is  intact,  and 
in  tension  where  the  bond  is  destroyed,  as  it  is  near  the  cracks. 
As  the  coefficient  of  contraction  during  hardening,  is  uncertain, 
the  same  analysis  as  above  cannot  be  employed,  but  the  whole 
amount  of  reinforcement  necessary  may  be  determined.  As 
before,  the  unit  tensile  and  bond  strengths  may  be  assumed 
to  be  practically  the  same,  and,  for  equilibrium,  the  strength 
of  the  concrete  cross-section  will  equal  that  of  the  steel.  The 
average  stress  in  the  concrete  will  be  half  the  ultimate,  as  it 
varies  from  that  amount  to  zero  after  the  cracks  occur.  Before 
the  concrete  is  thoroughly  set  the  tensile  strength  will  be  small, 
perhaps  not  over  150  Ib.  sq.  in.,  and  the  elastic  limit  of  the  steel 
may  be  taken  as  33,000  Ib.  sq.  in.  as  above.  Then  \  X  150  A  = 
pA  X  33,000  and  p  =  .0023.  As  there  is  much  uncertainty 
concerning  the  strength  of  concrete  during  this  stage  of  harden- 
ing this  result  is  only  approximate. 

Working  Rule.  It  is  the  common  custom  to  insert  about 
J  of  one  per  cent,  of  reinforcement  near  the  surface  of  a  wall 
to  provide  for  cracks  due  to  temperature  and  to  hardening, 
using  small  rods  for  the  purpose.  If  the  structure  be  fixed  in 
two  directions  the  reinforcement  must  be  placed  accordingly. 

PROBLEMS 

37.  A  simple  beam  of  20  feet  length  is  12  inches  wide  and  24  inches 
deep.     It  carries,  including  its  own  weight,  a  load  of  1200  Ib.  per  linear 
foot.     Find  the  size  and  spacing  of  the  stirrups,  allowing  30  Ib.  sq.  in. 
as  the  maximum  shearing  stress  in  the  concrete. 

38.  If  the  above  beam  be  reinforced  with  four  yf  inch  square  bars, 
at  what  point  may  the  two  inside  ones  be  bent  up? 

39.  What  uniform  load  will  the  beam  in  the  last  problem  carry  safely 
without  stirrups? 

40.  Prove  the  formulas  for  transverse  spacing  of  reinforcing  rods 
given  on  pages  82  and  83. 

41.  How  will  Fig.  41  be  changed  if  the  tension  in  the  steel  be  assumed 
to  decrease  gradually  from  the  elastic  limit  at  the  crack  to  zero,  where 
the  bond  is  perfect? 

42.  Prove  that  vertical  or  diagonal  reinforcement  tends  to  diminish 
the  tension  in  the  lower  horizontal  steel. 


90 


REINFORCED  CONCRETE 


FIG.  43. 


Combined  Flexural  and  Axial  Stresses  in  Beams.     Such  com- 
binations  of   stresses   may   occur  in  beams   confined   between 

walls,  in  columns,  dams,  retain- 
ing walls,  and  arch  rings.  The 
bending  moment  may  result  from 
lateral  pressure  or  it  may  be 
due  to  the  eccentric  application 
of  the  stresses  in  the  direction 
of  the  axis. 

While  not  practically  true, 
theoretically  the  same  condition, 
of  stiffness  in  a  section  may  be 
secured  by  substituting  for  the 
reinforcing  steel  n  times  its  area  of  concrete  at  the  same  dis- 
tance from  the  neutral  axis  as  was  the  steel.  Fig.  43  shows  the 
concrete  added  to  one  side  forming  what  is  known  as  the  trans- 
formed section.  The  area  of  the  transformed  section  is 

At  =  Ac  +  (n  -  1)  (A  +  A') 
in  which  Ac  is  the  area  of  the  concrete  section  bh, 
I,  =  I  +  (n  -  1)  /., 

in  which  It,  7,  and  Is  are  the  moments  of  inertia  of  the  trans- 
formed section,  of  the  concrete  section  bh,  and  of  the  steel  re- 
spectively, about  the  gravity  axis  of  the  concrete  section. 

If  the  section  be  rectangular  the  distance  from  the  upper 
fiber  to  the  center  of  gravity  is 


(74) 
(75) 

(76) 


+  (n  -  1)  /b  (Ad  +  A'd') 
+  (n-  1)  /b  (A  +  A') 

and  /  =  6/3  [u*  +  (h  -  u)*]. 

Since  p3  is  small, 

7s  =  A  (d  -  uY  +  A'  (u  -  d'Y  nearly. 


If,  as  in  (a),  Fig.  44  the  resultant,  R,  of  forces  on  one  side  of 
a  section  pierce  the  section  at  a  point  outside  the  gravity  axis, 
bending  will  result.  The  resultant  may  be  resolved  into  forces 
perpendicular  and  parallel  to  the  section,  as  H  and  V.  The 
former  is  the  normal  force  causing  flexure,  and  the  latter  is  the 


ANALYSIS  OF  STRESSES 


91 


shear.     If  R  be  applied  a  distance,  e,  from  the  gravity  axis,  the 
bending  moment  is 

M  =  He 


and 

also 


C  = 

C'  = 


H 


Ac  +  (n  -  1)  (A 
H 


A') 


+ 


He-u 


+  (n-  1)  7, 

He  (h  -  u) 

Ac  +  (n  -  1)  (A  +  A')  ~  I  +  (n-  1)  /s 


The  maximum  or  minimum  fiber  compression  is,  thus,  due 
to  the  uniform  load,  H,  over  the  entire  section  and  to  the  bend- 
ing. In  Fig.  44  the 

i  H — C ^ 

uniform  load  alone 
would  be  rectangu- 
lar and  of  the  width 
shown  on  xx,  the 
gravity  axis.  The 
difference  between  -i- 
such  a  diagram 
and  the  one  shown 
would  be  the  stress 
diagram  for  bending 
alone. 


FIG.  44. 


The  unit  stress  in  the  steel  is  n  times  that  in  the  concrete 
fibers  at  the  same  distance  from  the  axis;  so  if  the  steel  be  located 
as  shown  in  the  figure, 


LC  +  (n  -  1)  (A  +  A') 


*«  fa  ~  dO  .1     (79) 
/  +  (n  -  1)  /J 


and 


H £fctf-u)_-j 

Ac  +  (n  -  1)  (A  +  A')       I  +  (n  -  1)  /J 


where  $c  is  the  stress  in  the  steel  on  the  same  side  of  the  axis  as 
H,  and  $'c  in  (a),  and  $<  in  (6),  refer  to  steel  on  the  opposite  side. 
These  formulas  are  general,  whatever  the  form  of  the  cross- 
section,  and  whatever  the  eccentricity  may  be,  and  are  readily 
applied  when  numerical  values  of  the  factors  are  substituted. 
When  the  eccentricity  is  so  great  that  concrete  in  tension  becomes 
cracked,  the  effective  area  of  the  concrete  is  no  longer  bh,  while 


92  REINFORCED  CONCRETE 

/  and  u  become  variable.  In  this  case  the  above  formulas  are 
not  readily  applied  and  others  will  be  deduced  later.  This 
case  occurs  when  the  value  of  C'  in  (78)  becomes  negative  and 
will  be  called  Case  II. 

Case  I.  Rectangular  Section  with  Symmetrical  Reinforce- 
ment, all  in  Compression.  In  very  many  instances  the  section 
considered  is  rectangular,  and  the  above  formulas  may  be  made 
more  simple  when  numerical  values  are  substituted.  In  designs 
wherein  compression  extends  over  the  entire  section,  the  rein- 
forcement is  frequently  placed,  in  equal  amounts,  at  the  same 
distances  from  the  outer  fibers  on  either  side.  Here  the  trans- 
formed section  is  symmetrical  about  the  axis,  midway  between 
the  top  and  bottom,  u  =  \h  and  /  may  be  taken  as  yV  bh3. 
As  already  defined,  2  p  is  the  area  of  the  steel  divided  by  the 
effective  area  of  the  whole  section,  so  2  p  =  (A  +  A')  -f-  bh. 
Making  these  substitutions 

c         H   f  1  Qeh \ 

"  bh  VI  +  2  (n  -  1)  p  ^  h2  +  24  p  (n  -  1)  (u  -  d')2) 

c,   .-  H  (  l ^L ^     (*2] 

~  bh  VI  -f  2  (n  -  1)  p      h2  +  24  p  (n  -  1)  (u  -  d')2) 

1  ,  12  (u  -  d'}  e \ 

.1  H-  2  (n  -  1)  p  ^  h2  +  24  p  (n  -  1)  (u  -  d')2) 


bh  U  +  2  (n  -  1)  p      h2  +  24 


2^dOj_          \ 
p  (n  -  1)  (t*  -  W)*J 


If  the  value  of  e  be  such  that  C'  is  more  than  the  safe  stress  for 
concrete  in  tension,  formula  (82)  does  not  apply,  and  the  com- 
putation must  be  made  under  Case  II. 

Diagram  for  Case  I.     If  values  be  assumed,  for  n  =  15  and 
for  u  —  d'  =  0.4  h,  the  formulas  become 


C 
C' 

-3.' 

•  H( 

f 

i 

+  h 
e 

6 

W  =  A* 

H             (SVi 

U 

f 

+  28p 

1 

I 

+ 

53.8 
6 

'  bh 

''bh    ^ 

bh\ 

a 

/ 

+  28p 
15 

h 

1 

+ 

53.8 

79 

P) 
\ 

Sc 

r 

_i_  e 

sH 

"•  bk    w 

~    bh 

(i 

+28  p 

1   h 

1 

+ 

53. 

8 

ANALYSIS  OF  STRESSES 


93 


PLATE  V1I1. 


94 


REINFORCED  CONCRETE 


Sc' 


15 


72 


bh  \l  +  28  p       h  1  +  53J 


-  "-"'a 


(88) 


where  p  is  half  the  total  amount  of  steel  in  the  whole  section. 

Values  of  C  and  C"  are  plotted  on  Plate  VIII,  and  of  Sc  and 
S'c  on  Plate  IX,  the  diagrams  extending  only  over  such  values 
of  e  -T-  h  as  being  the  problem  under  Case  I.  It  is  apparent 
that  C  is  the  critical  value,  as  it  is  always  greater  than  C',  and 
also  greater  than  Sc/n.  In  the  diagram  p  is,  for  convenience, 
the  proportion  of  steel  at  each  edge. 


i 

* 

5? 

3 

; 

^* 

Values 

ofp  +  p1 

s^ 

Case  II 

// 

X 

ft   fW\    '  1* 

^^ 

-j.  ()  ->Q 

5 

X' 

FLlQi-8 

X 

(-low 

X 

/ 

Case  I 

X 

Values 

ofp  +  p 

X 

i 

* 

2 

* 

8 

% 

FIG.  45. 

Fig.  45  shows  the  curve  separating  Case  I  from  Case  II. 
For  example,  if  p  +  pf  be  \  of  1  per  cent.,  values  of  e  -f-  /i  under 
.177  indicate  Case  I,  and  larger  values  Case  II. 

If  the  rectangular  section  be  that  of  a  column,  free  to  bend 
in  any  direction,  the  denominator  in  e/h  is  not  necessarily  the 
longer  of  the  two  axes,  but  the  fraction  must,  for  safe  design, 
be  the  larger  of  the  two  values,  e/h  and  e/b.  In  the  case  of  a 
beam,  eccentricity  in  the  direction  of  the  shorter  dimension  of 
the  section  is  not  usually  considered,  since  a  floor  or  other  brac- 
ing is  usually  attached  in  such  manner  as  will  prevent  lateral 
flexure. 

Case  II.  Tension  and  Compression  in  the  Section.  The 
formulas  in  Case  I  apply  when  there  are  small  tensile  stresses 
in  the  concrete  such  as  will  not  cause  cracks.  When  the  latter 


ANALYSIS  OF  STRESSES 


95 


0.002 


20- 


\ 


).OOG 


0.01          |  0.002 

[Values  of  p=p' 
Total  Steel  =(p  -f p)  Wi 


-14- 


\ 


\ 


\ 


-10 


\ 


d 

, 


PLATE  IX. 


96 


REINFORCED  CONCRETE 


fraction  within  the  parenthesis  of  (82)  becomes  larger  than 
the  first,  C"  changes  sign  and  the  numerical  value  indicates  the 
unit  tension  in  the  concrete. 

In  Fig.  46  all  the  tension  is  assumed  to  be  in  the  steel.     Then 
for  equilibrium  in  a  rectangular  section 


(89) 


8,.  ^  n 


'       k 
M-§. 

the  center  of  moments  being  at 
the  gravity  axis  of  the  section. 

Because  of  the  conservation  of 
plane  sections 

C:St  +  n  =  k:d  -  k 
C:SC  -T-  n  =  k:k  -  d' 

These  equations  may  be  solved  when 
numerical  values  are  inserted,  but  the  lit- 
eral expression  for  k  is  quite  complicated. 
When  p  =  p'  and  d'  =  h  —  d,  u  becomes 
h/2,  and  the  expressions  may  be  made  more 
simple.  With  some  approximations,  a  for- 
mula involving  k  may  be  written  after  multiplying  (89)  by  e 
and  equating  it  to  (90). 


1. 


fc3 


-  3  k2  Q  -  e\  +  12  kenph  -  6  pn  (eh2  +  2  ufh)  =  0 


(91) 


where  Ui  is  h/2  —  d'  and  e  is  the  eccentricity  of  the  point 
of  application  of  the  external  forces.  The  value  of  k  may  be 
found  by  trial,  or,  when  n  =  15  and  u\  =  &  h,  in  Plate  X. 
After  k  is  found  the  stresses  may  be  determined  for  given  values 

of  H  or  M  and  e. 

2  Hk 


St  =  nC 


bk*  +  2  pbhn  (2k  -h) 
+  h  -2k 


2k 


(93) 


Sc  =  nC 


2k 


(94) 


ANALYSIS  OF  STRESSES 


97 


PLATE  X. 


98  REINFORCED  CONCRETE 

If  the  reinforcement  be  placed  h/10  from  the  upper  and  lower 
edges,  pbh  at  each  point,  HI  =  A  h,  and  the  last  three  formulas 
reduce  as  follows: 

2  Hk 

"  bk*  +  30  pbh  (2k  -  h) 

Cr  =  St  =  15  C  '9  h~  k  (96) 


Cr'  =  Sc  =  15  C    -~-  (97) 

.     r  =  ^o^^          *  (98) 

also  M  =  e-h  (J^  +  30  p  -  15  &f\  Cbh*  (99) 

=  NiCbh2  (100) 

In  these  formulas  p  is  the  reinforcement  at  both  top  and 
bottom  of  the  beam,  and  is,  hence,  half  the  whole.  Also,  since 
these  equations  and  those  of  Case  I  are  intimately  connected, 
p  is  A  -f-  bh  rather  than  A  -f-  bd.  In  plotting  values  of  A:  in 
Plate  X,  or  in  extending  the  same  if  desired,  it  is  much  simpler 
to  assume  k  and  solve  for  e/h  since  this  factor  is  less  involved 
than  is  k. 

For  example,  to  illustrate  the  use  of  Plates  VIII-XI  and  the 
formulas  upon  which  they  depend,  let  the  depth  of  an  arch 
ring  be  15  inches,  and  let  the  resulting  pressure  be  6000  pounds 
for  each  inch  in  length.  If  the  reinforcement  be  .0075  at  h/10 
from  intrados  and  extrados,  and  the  resultant  pressure  be  applied 
2  inches  from  the  middle,  and  at  an  angle  of  5°  with  the  right 
section,  what  will  be  the  unit  pressure  in  the  concrete  at  each 
edge  and  in  the  inner  and  outer  steel?  Here  e/h  is  2  -f-  15  = 
0.133,  and,  by  Fig.  45,  the  problem  falls  clearly  in  Case  I.  The 
normal  pressure,  H,  for  1  inch  of  breadth  is  6000  X  cos  5°  = 
5976,  and  the  moment  is  5976  X  2  =  11,952  Ib.  in.  By  formula 
(85),  or  by  the  diagram  in  Plate  VIII, 

5976 


c 


1  X  15 


/I  J3  __  \ 

.  \1  +  .14  XI."  1  +  .269  X  {/ 


X  1.39  =  560  Ib.  sq.  in,  C'  =  X  .256  =  104  Ib.  sq.  in. 


.  .     .     , 

15  A  X 


ANALYSIS  OF  STRESSES 


99 


PLATE  XI. 


100  REINFORCED  CONCRETE 

In  the  lower  part  of  Plate  VIII,  the  factor  .256  is  taken  directly. 
„        5976  /  15          9.6 


and  S'e  =    nr  X  5-6  =  223°  lb-  sq-  in- 

lo 

The  factors  for  the  stress  in  the  steel  are  found  in  Plate  IX. 

Let  the  following  assumptions  be.  made:  6  =  8  inches,  h  =  35 
inches,  A  =  A'  =  1.12  square  inches,  M  =  520,000  lb.  in., 
n  =  15,  and  reinforcement  placed  symmetrically.  If  the  mo- 
ment be  caused  by  an  axial  load  of  29,000  pounds,  it  is  required 
to  find  the  stresses  in  the  concrete  and  in  the  steel.  The  per- 
centage of  reinforcement  is  .4,  and  the  eccentricity  is  520,000  -r- 
29,000  =  17.9  inches.  A  reference  to  Fig.  45  shows  that  the 
problem  belongs  to  Case  II  and  in  Plate  XI,  for  p  =  0.4  per 
cent,  and  e/h  =  0.512,  N7  is  found  to  be  0.122.  Then  C  = 
520,000  -s-  (.122  X  8  X  352)  =  433  lb.  sq.  in.  At  the  bottom 
of  Plate  XI,  r  is  found  to  be  14,  and  from  (97)  rf  is  11.7,  so 
St-  =  433  X  14  =  6060,  and  Sc  =  433  X  11.7  =  5100  lb.  sq.  in. 

A  beam  is  14  inches  wide  and  24  inches  deep.  The  bending 
moment  is  525,000  \b.  in.,  and  the  eccentricity  is  8  inches.  If 
the  working  strength  of  the  concrete  be  600  lb.  sq.  in.,  how 
should  the  beam  be  reinforced.  Here  the  eccentricity  ratio  is 
8  -r-  24,  and  Fig.  45  shows  the  problem  to  be  one  for  the  appli- 
cation of  Case  II  for  any  value  of  p  there  given.  In  Plate  XI, 
N7  =  525,000  -T-  (600  X  15  X  24  X  24)  =  0.109.  At  the  inter- 
section of  this  value  of  Ni  and  e/h  =  0.3,  p  is  seen  to  be  0.24 
per  cent.,  which  is  also  the  value  of  p'. 

In  the  last  example  let  e/h  be  changed  to  J,  involving  Case  I, 
as  shown  by  Fig.  45.  Ni  =  600  X  24  X  14  X  24  -f-  525,000  = 
1.54.  In  Plate  VIII  this  factor,  with  e/h  =  J,  calls  for  about 
0.74  per  cent,  of  steel  at  each  edge.  It  will  be  noted  that  this 
problem  assumes  the  same  bending  moment  as  the  last,  while 
the  eccentricity  has  been  reduced.  This,  of  course,  means  that 
the  axial  force,  H,  must  be  increased  in  the  same  ratio,  and  the 
necessary  increase  in  the  percentage  is  thus  explained. 

PROBLEMS 

43.  Read,  in  Transactions  of  American  Society  of  Civil  Engineers,  Vol. 
XLI,  the  article  on  safe  stresses  in  steel  columns. 


ANALYSIS  OF  STRESSES  101 

44.  Deduce  a  formula  for  M  =  NCbh2  in  Case  I,  by  the  method 
employed  in  Case  II. 

45.  The  bending  moment  in  a  certain  section  of  a  beam  is  500,000 
Ib.  in.,  the  resulting  stress  being  applied  4  inches  from  the  middle,  the 
breadth  and  depth  are  12  and  20  inches  respectively.     If  the  reinforce- 
ment be  p  =  p'  =  .8  per  cent.,  what  values  have  C,  C',  Sc,  and  &? 

46.  In  the  last  problem,  what  will  be  the  stresses  if  the  same  axial 
force  be  applied  2  inches  from  the  middle? 

47.  In  the  same  problem,  what  will  be  the  stresses  if  the  same  bending 
moment  be  occasioned  by  an  axial  force  applied  2  inches  from  the  middle? 

48.  Design  a  reinforced  concrete  beam  of  1:2:4  concrete  and  mild 
steel,  the  length  being  18  feet,  the  load  500  pounds  per  linear  foot,  besides 
the  weight  of  the  beam,  and  the  axial  load  2000  pounds,  applied  4  inches 
from  the  middle  of  the  section. 

Flexure  and  Axial  Tension.  Whether  under  Case  I  or  Case  II, 
the  above  discussion  applies  as  well  to  tension  as  to  compression, 
when  either  stress  is  added  to  that  of  bending.  In  (81)  and 
(82)  it  is  to  be  noted  that  both  C  and  C'  depend  upon  the  sum 
of  the  stresses  due  to  direct  compression  and  to  flexure.  The 
first  fraction  in  these  equations  give  the  compressive  stress  due 
to  axial  forces,  and  these  stresses  will  simply  change  sign  if  the 
force  parallel  to  the  axis  be  changed.  Hence,  the  same  formulas 
and  diagrams  apply  for  axial  tension  as  for  axial  compression 
under  Case  I. 

The  formula  used  in  finding  the  value  of  k  (74)  applies  when 
the  direction  of  the  thrust  is  considered.  Since  He  =  M,  the 
sign  of  e  changes  with  that  of  H,  so  the  formula  for  K  in  the 
case  of  axial  compression  is  changed  to  suit  the  case  of  axial 
tension  if  the  sign  of  e  in  (91)  be  changed  wherever  it  occurs. 
The  formula  then  becomes 

fc3  -  3  k2  (h/2  +  e)  -  12  kenph  +  6  pn  (eh2  -  2  ufh)  =  0 

-2Hk 
bk2  +  2  pbhn  (2k  -h) 

With  these  formulas  problems  of  this  nature  may  be  solved  or 
diagrams  similar  to  Plates  IX  and  X  may  be  readily  constructed. 
Instances  of  such  a  combination  of  stresses  are  rather  rare 
in  reinforced  concrete  design,  since  masonry  of  any  kind  is  not 
well  adapted  to  take  tension;  but  in  any  case  there  should  be 
no  trouble  experienced  in  adapting  any  formula  for  a  positive 
thrust  to  a  problem  involving  thrust  in  an  opposite  direction. 


102  REINFORCED  CONCRETE 

PROBLEM 

49.  Construct  diagrams  similar  to  Plates  X  and  XI  for  values  of  k 
and  N'7  when  axial  tension  is  combined  with  flexure. 

Columns 

Concrete  is  widely  used  for  columns,  and  may  or  may  not 
.be  employed  with  steel  in  this  way.  When  steel  is  used  in  this 
connection,  the  concrete  is  sometimes  but  a  covering  or  filler 
for  the  purpose  of  adding  beauty,  symmetry,  and  fire  protection. 
It  is  then  not  essential  to  the  security  of  the  structure  although 
it  gives  some  additional  strength  and  stiffness.  This  type  is 
not  strictly  reinforced  concrete,  and  the  analysis  of  stresses 
in  such  designs  need  not  be  taken  up  here.  The  steel  in  these 
columns  is  made  up  of  structural  shapes  in  a  variety  of  designs, 
and  no  part  of  the  load  is  assigned  to  the  concrete.  This  style 
of  column  is  effective,  and  is  to  be  highly  commended.1 

Plain  columns  will  be  referred  to  for  the  particular  purpose  of 
comparing  the  strength  of  such  with  that  of  columns  reinforced 
in  different  ways.  It  may  be  said  that  the  most  economical 
reinforcement,  or  means  of  strengthening,  is  additional  or  better 
cement  in  the  mortar  of  which  the  column  is  composed.  Where 
the  space  is  not  too  limited,  plain  concrete  is  much  used  for 
short  columns. 

Columns  with  Longitudinal  Reinforcement.  In  general,  col- 
umn reinforcement  is  of  two  kinds,  longitudinal  and  circum- 
ferential. The  latter  kind  is  usually  in  the  form  of  a  spiral, 
either  flat  or  of  wire,  or  made  of  bands  or  hoops,  and  is  called 
"  hooped."  The  intended  action  of  these  two  varieties  of 
reinforcement  is  very  dissimilar  in  the  two  cases.  In  the  former 
the  metal  acts  in  direct  compression,  and  its  shortening  under 
the  load  is  the  same  as  that  for  the  concrete,  the  accompanying 
stresses  depending  upon  the  coefficient  of  elasticity  of  the  steel 
and  of  the  concrete.  In  the  hooped  columns  the  metal  is  used 
to  confine  the  concrete,  which  takes  all  the  axial  stress,  thus 
preventing  disintegration  by  crushing  or  by  diagonal  shear. 

Upon  the  assumption  that  the  load  is  uniformly  distributed 
over  the  cross-section  of  the  column,  the  steel  should  be  sym- 

1  For  an  example  of  the  use  of  structural  steel  with  concrete  see  Trans. 
Am.  Soc.  of  C.  E.,  1908,  Vol.  LX. 


ANALYSIS  OF  STRESSES 


103 


metrically  distributed  over  that  area.  The  distance  of  the  metal 
from  the  outside  would  be  immaterial,  and  the  reinforcement 
might  even  be  placed  in  the  middle.  However,  columns  are 
never  exactly  straight,  and  the  loads  are  seldom  centrally  applied, 
so  the  rods  are  usually  not  over  2  inches  from  the  outside.  Fig. 
47  (6)  shows  bands  of  wire  used  to  prevent  buckling  of  the  rods 


under  excessive  loads,  when  the  concrete  takes  a  continually 
decreasing  proportion  of  the  load.  Since  the  deformation  of 
steel  and  concrete  is  the  same,  the  stresses  vary  with  Ec  and 
Es',  the  value  of  n  may  change  from  12  to  25  during  a  test  of  a 
column  and  the  ratio  of  stresses  changes  accordingly. 


Let 
Then 

Solving, 
also 


A  =  the  area  of  steel  in  the  section 
Ac  =  the  total  area  of  the  section 

P  =  the  total  load  on  the  column. 

P  =  C  (Ac  -  A)  +  SCA 
=  C  (Ac  -  pAc  +  npAc) 
=  CAC  [(I  +  p(n-  1)] 

P  -  CAC 


P  = 

Ac  = 


ACc  (n  -  1) 

P 

C[l+p(n-  1)] 


(101) 
(102) 

(103) 


The  excess  of  strength  of  the  reinforced  over  the  plain  concrete 
column  is  ,  /rl 

A  (Se  -  C)  =  PACC  (n  -  1)  (104) 

and  the  ratio  of  increase  is 

pAcC  (n  -  1) 

-  =p(n"  J> 


The  ratio  of  increase  in  strength  is  seen  to  be  independent  of  C. 


104 


REINFORCED  CONCRETE 


These  relations  are  based  upon  the  supposition  that  the  rods 
and  the  concrete  are  firmly  connected  by  the  bond,  so  that  they 
act  together.  Experiments  show  that,  under  ordinary  condi- 
tions, this  assumption  is  correct,  and  that  the  steel  will  carry 
the  proportion  of  the  working  load  indicated  by  the  above 
formulas.  Ultimate  failure  is  sometimes  caused  by  the  buck- 
ling of  the  reinforcement  which  causes  the  outside  concrete  to 

scale  and  the  cross- 
section  to  be  there- 
by reduced.  Hence 
the  steel  should  not 
be  placed  too  near 
the  surface  and 
about  1J  inches  of 
concrete  usually 
covers  the  rods  on 
the  outside. 

As  an  example 
of  the  use  of  these 
formulas,  let  it  be 
required  to  deter- 
mine the  proper 
size  of  a  square 
column,  reinforced 
with  2  per  cent,  of 
steel,  to  carry  a 
load  of  100,000 
pounds  with  a  unit 
stress  of  450  pounds  in  the  concrete.  Let  n  =  15.  In  (103) 
Ac  =  100,000  -r-  450  (1  +  0.02  X  14)  =  174  square  inches,  or 
the  column  should  be  13.2  inches  square. 

Again,  what  will  be  additional  strength  of  the  column  if  the 
reinforcement  be  increased  to  2J  per  cent.?  From  (101)  the 
increase  in  strength  changes  from  0.02  X  14  X  174  X  450  to 
.025  X  14  X  174  X  450  or  10,960  pounds  for  each  percentage  of 
steel.  In  the  first  case  the  strength  of  the  concrete  column  has 
been  increased  by  0.02  X  14  or  28  per  cent.,  and  in  the  second 
case  by  0.025  X  14  or  35  per  cent,  by  the  addition  of  the  steel. 

These  relations  are  shown  in  Fig.  48.  The  disproportion 
between  the  strength  of  steel  and  that  of  concrete  is  emphasized 


Strengih  of  Reinforced  to  Plain  Columns 
Ul  JJ2  1J3  14 


FIG.  48. 


ANALYSIS  OF  STRESSES  105 

more  as  the  latter  becomes  leaner,  or  as  n  becomes  greater. 
Hence,  for  the  same  percentage  of  steel,  the  percentage  of  in- 
crease in  strength  of  the  reinforced  over  the  plain  column  is 
greater  with  the  leaner  concrete.  Referring  to  Fig.  48,  it  is 
seen  that  2  per  cent,  of  steel  increases  the  strength  of  a  plain 
column  38  per  cent,  when  E  =  1,500,000,  but  the  increase  is 
only  18  per  cent,  if  E  be  3,000,000,  or  if  n  be  10. 

As  an  example  of  the  use  of  Fig.  48,  let  the  column  be  1 1  X  11 
inches,  the  load  60,000  pounds,  and  the  working  strength  of 
the  concrete  be  400  Ib.  per  sq.  in.  If  n  be  15,  what  per  cent, 
of  steel  is  needed?  The  stress  per  square  inch  of  cross-section 
is  60,000  -r-  121  =  496  pounds,  while  but  400  pounds  is  to  be 
allowed  per  square  inch  of  concrete.  So  the  per  cent,  of  increase 
in  strength  must  be  24,  and  the  diagram  shows  that  1.7  per  cent, 
of  steel  is  needed. 

Tests  of  Plain  Columns.  The  following  table  is  a  summary 
of  tests  of  plain  columns  taken  from  Bulletin  No.  20,  of  the 
University  of  Illinois  Engineering  Experiment  Station,  1907. 
These  results  should  be  compared  with  those  for  compression 
of  cubes  given  on  page  16.  These  columns  were  cylindrical, 
12  inches  in  diameter,  and  10  feet  long.  In  general,  failure  took 
place  in  the  middle  half  of  the  length,  the  exceptions  being 
when  local  imperfections  were  apparent. 

"In  general,  two  somewhat  distinct  forms  of  failure  were  observed, 
(a)  a  failure  which  may  be  termed  a  diagonal  shearing  failure,  and  (6)  a 
failure  which  may  be  termed  simple  compression.  In  the  first  the  fracture 
was  angular  in  nature,  having  the  appearance  of  a  diagonal  shearing 
failure  characteristic  of  the  manner  of  failure  in  compression  so  common 
in  brittle  materials.  In  these  failures  the  columns  broke  suddenly  and 
without  warning,  some  of  them  breaking  after  the  machine  had  been 
stopped  and  while  no  additional  load  was  being  applied.  A  loud  report 
accompanied  the  failure.  The  second  column  failed  suddenly  and  had 
given  so  little  warning  that  the  instruments  had  not  been  removed.  In 
the  second  form  of  failure,  here  called  "  simple  compression,"  the  column 
shattered,  cracking  longitudinally  for  some  distance,  the  cracks  being 
well  distributed  over  the  faces  of  the  column.  In  some  of  these,  the  failure 
was  not  noted  until  the  weighed  load  began  to  decrease,  and  the  position 
of  the  failure  was  not  determined  until  the  machine  had  produced  a  fur- 
ther shortening  of  the  column.  As  will  be  seen  in  the  table,  nearly 
all  the  richer  mixtures  (1-1 £-3  and  1-2-4)  gave  diagonal  shearing 
failures,  and  the  columns  made  with  lean  mixtures  crushed  throughout 
the  whole  fracture  in  what  are  here  termed  simple  compression  failures. 


10t3 


REINFORCED  CONCRETE 


.  .  .  For  both  fonns  of  failure,  it  may  be  stated  in  general  that  the 
approach  of  the  ultimate  strength  of  the  column  might  have  been  pre- 
dicted from  the  increase  in  the  rate  of  shortening  and  from  the  shape  of 
the  load-deformation  diagram." l 

SUMMARY  OF  TESTS  OF  PLAIN  CONCRETE  COLUMNS 


Mixture 

days 

i«!STte^ 

peraq.  in. 

Manner  of  failure 

Initial  mod- 
ulus of 
elasticity 

1-11-3 

66 

2120 

Diagonal  shearing  without  warning     2,800,000 

« 

62 

M80 

u                     u                   u                   ft 

3.890,000 

1-2-4 

58 

1165 

Diagonal  shearing 

2^30,000 

u 

60 

mo 

General  crushing 

3,350,000 

a 

65 

2210 

u 

3.440,000 

u 

64 

1590 

U                            It 

3,390,000 

tf 

62 

1945' 

Diagonal  shearing 

2,660,000 

u 

72 

1460  * 

General  crushing 

3,380,000 

u   ' 
1    l-ft 

M 

MI 

1810 

QrlS 

Sadden  diagonal  shearing 

2,830,000 

1—  «>-O 
u 

1    —1       X 

Dl 

62 

K3. 

W99 

1110 

=7= 

General  crushing 

— 

1   *»-o 

(I 

Oo 

63 

Of  O 

575 

MM                           it 



1-2-4 

203 

1925 

General  crushing 

3,270,000 

u 

194 

1845 

*<              t< 

3,140,000 

u 

181 

1770 

n             n 

3,600,000 

u 

187 

9880 

Sudden  diagonal  shearing 

3,000,000 

u 

187 

2160 

Diagonal  shearing 

3,100,000 

u 

201 

1770 

General  crushing 

2,770,000 

1-2-31 

12mos, 

aeso 

Diagonal  shearing 

— 

ft 

16mos. 

2770 

General  crushing 

~ 

The  table  shows  clearly  that  the  richer  mixtures  produce 
columns  of  much  higher  strength  than  do  the  lean  ones.  The 
average  strength  of  the  several  mixtures  is  as  follows:  1-1 J-3 
concrete,  2300  Ib.  per  sq.  in.;  1-2-4  concrete,  1740  Ib.  per 
sq.  in.;  1-3-6  concrete,  1030  Ib.  per  sq.  in.,  and  1-4-8  con- 
crete, 575  Ib.  per  sq.  in. 

Working  Load  for  Plain  Concrete  Columns.  The  safe-unit 
load  for  such  columns  as  recommended  by  the  Joint  Committee 
(see  page  152)  is  450  Ib.  per  sq.  in.  on  concrete,  of  which  cubes 
will  hold  2000  Ib.  sq.  in.  The  length  of  the  columns  should 
not  exceed  12  diameters,  the  load  should  not  be  eccentric,  and 

1  Made  of  Universal  cement.     All  others  made  of  Chicago  A  A. 
*  Quotation  from  above-named  bulletin. 


ANALYSIS   OF  STRESSES 


107 


the  age  should  be  28  days,  otherwise  the  stated  safe  load  should 
be  modified  accordingly. 

Tests   of   Columns   with  Longitudinal   Reinforcement.     The 
following  table  is  from  Bulletin  No.  10,  University  of  Illinois 
Engineering  Experiment  Station.     The  mixture  was,  in  each 
case,  l-2-3|  concrete.     The  elastic  limit  of  the  steel  was  39,800, 
pounds  per  square  inch. 


Ln«dl 

feet 

sz 

Gran- 

"sss 

Reinforcement 

Crushing 

-r-^:r.  :-. 
peraq.in. 

Modulus 
of 
elasticity 

Variety 

PrT  '/rZL* 

12 

71 

12  X12 

4—  f  in.  rods 

1.20 

1587 

2,570,000 

12 

71 

It 

54-|"      " 

1.21 

1862 

2,340,000 

i  12-|  "   ties 

12 

65 

ft 

4-i  "   rods 

1.21 

1850 

2,570,000 

12 

65 

tt 

5    4-f  "      « 

1.21 

1936 

2,430,000 

t  12-1  "    ties 

12 

69 

9X9 

4-|"   rods 

1.52 

1577 

2,330,000 

12 

70 

u 

4-|  "      " 

1.52 

1600 

2,090,000 

12 

65 

H 

5    4-|  "      « 

1.50 

1280 

1,800,000 

1  12-1  "    ties 

9 

66 

M 

j    4-f  "    rods 

1.48 

2335 

2,500,000 

*    9-i  "    ties 

12 

63 

it 

J    4-|"    rods 

1.50 

1367 

2,000,000 

t  12-i  "    ties 

9 

59 

u 

(    4-|  "    rods 

1.49 

1607 

1,900,000 

*    9-i  "    ties 

6 

67 

tt 

4-|  "    rods 

1.47 

2206 

1,900,000 

12 

69 

12X12 

none 

0 

1710 

3,150,000 

12 

64 

9X9 

it 

0 

2004 

2,530,000 

12 

65 

12  X12 

u 

0 

1610 

2,500,000 

12 

61 

u 

n 

0 

1709 

2,370,000 

6 

63 

It 

u 

0 

1189 

2,000,000 

6 

65 

9X9 

tt 

0 

1079 

1,490,000 

When  the  concrete  is  stressed  to  the  ultimate,  n  is  probably 
as  much  as  18.  Using  this  value  in  (103),  and  substituting  the 
average  values  for  the  12  inches  X  12  inches  —  12  feet  columns 
with  1.21  per  cent,  reinforcement,  gives 

C  (1  +  .0121  X  17)  =  1809 

C  =  1500  Ib.  per  sq.  in. 

The  average  strength  of  the  plain  columns  of  the  same  dimen- 
sions is  1676  Ib.  per  sq.  in.  For  the  9"  X  9"-12/  columns  the 


108 


REINFORCED  CONCRETE 


computed  value  of  C  is  1160,  and  the  table  gives  2004  Ib.  sq.  in. 
for  a  single  test,  showing  that  the  assumed  value  of  n  is  much 
too  large  for  agreement.  The  result  corresponds  fairly  well 


FIG.  49a. 


FIG.  496 


with  the  minimum  crushing  strength  for  plain  concrete  in  this 
table,  and  also  in  that  on  page  16  for  1:2:4  mixture. 

Ultimate  failure  of  columns  reinforced  in  this  way  sometimes 
takes  place  through  the  buckling  of  the  rods  between  the  ties 
which  are  usually  spaced  about  a  foot  apart. 


ANALYSIS  OF  STRESSES 


109 


Working  Load  for  Columns  with  Longitudinal  Reinforcement. 

As  given  on  page  152  the  recommended  unit  stress  over  the  cross- 
section  of  the  column  rein- 
forced only  with  longitudinal 
rods  is  450  (1  +  14  p)  Ib.  sq. 
in.  Since  the  value  of  Ec  be- 
comes materially  less  as  the 
unit  stresses  increase,  the  fac- 
tor of  safety  obtained  by  divid- 
ing the  ultimate  by  the  working 
stress  is  less  than  that  actually 
resulting  from  the  use  of  that 
working  stress.  Hence,  in  case 
the  ultimate  strength  of  a  par- 
ticular mixture  is  definitely 
known,  the  factor,  450,  in  the 
above  formula  may  be  changed 
to  one  found  by  dividing  the 
known  strength  by  a  suitable 
factor  of  safety,  usually  4. 

Columns  with  Hooped  Rein- 
forcement. In  this  construction 
the  office  of  the  reinforcement 
is  only  to  prevent  lateral  dis- 
placement of  the  component 
parts  of  the  concrete  while  the 
column  is  under  axial  stress. 
Foundations  of  sand  are  known 
to  be  stable  if  the  tendency  of 
the  grains  to  move  in  a  plane 
perpendicular  to  the  direction 
of  the  pressure  be  prevented. 
The  bands,  hoops,  or  spirals  do 
not  and  cannot  take  any  of  the 
stress  in  the  direction  of  the  axis  of  the  column,  but  resist  the 
lateral  component  of  the  diagonal  shear  and  the  swelling  shown 
in  Figs.  50  and  51. 

The  first  published  reports  of  experiments  on  columns  of  this 
kind  were  made  in  1902  by  Considere,  in  France.1     Later,  among 
1  Beton  und  Eisen,  No.  V,  1902.     Genie  Civil,  November,  1902. 


FIG.  50. 


110 


REINFORCED  CONCRETE 


other  forms,  there  were  columns  tested  of  octagonal  section  1  m. 
long  and  27.5  cm.  in  diameter  at  the  spiral.  The  reinforcement 
consisted  of  wire  wound,  with  various  spacing  or  pitch,  around 

longitudinal  rods,  arranged  sym- 
metrically in  a  circle  15  mm. 
from  the  outside.  These  tests 
were  made  at  Stuttgart  in  1905 
and  showed  decided  increase  in 
strength  of  such  columns  over 
those  of  plain  concrete,  and  the 
publication  of  the  results  cre- 
ated wide  interest  in  this  type 
of  construction.  Considere  con- 
cluded that  the  ultimate  load 
for  such  columns  was  1.5  A  \C 
+  Sf  (A2  +  24  At),  in  which  A^ 
is  the  area  of  the  cross-section 
inside  the  spiral,  assumed  to  be 
f  of  the  whole  area,  C  is  the  ulti- 
mate stress  of  the  unreinforced 
concrete,  $'  is  the  elastic  limit 
of  the  steel,  A2  is  the  area  of 
the  cross-section  of  the  rods, 
and  As  is  that  obtained  by 
dividing  the  volume  of  the 
spiral  by  the  volume  of  the 
column  inside  the  spiral.  This 
formula  assumes  that  the  spiral 
is  2.4  times  as  effective  as  the 
longitudinal  steel.  The  above 
equation  may  be  stated  in  gen- 
eral form  as 


FIG.  51. 


Ci  =  C  +  pm 

in  which  Ci  is  the  unit  strength 
of  the  column,  C  is  the  unit  strength  of  a  plain  column  of 
like  dimensions,  p  is  the  volume  of  the  spiral  divided  by  that 
of  the  column  core,  and  m  is  a  factor,  found  by  experiment, 
indicating  the  effect  of  the  hoops  upon  the  strength  of  the 
column. 

While  the  hooping  on  columns  is  of  great  value  in  making 


ANALYSIS  OF  STRESSES 


111 


them  capable  of  large  deformation  before  breaking,  it  may  be 
easily  shown  that  it  can  exert  very  little  influence  upon  the 
strength  of  the  column  until  the  ultimate  strength  of  the  con- 
crete has  been  reached. 

Most  substances,  when  subjected  to  tension  or  compression 
along  one  axis,  will,  as  a  consequence,  contract  or  expand  along 
axes  perpendicular  to  the  first.  The  quotient  obtained  by 
dividing  this  secondary  deformation  by  the  first  is  called  Pois- 
son's  ratio,  and  is  represented  by  u.  In  the  case  of  concrete 
under  working  stresses,  this  ratio  has  been  found  to  be  from 
xV  to  J. 

Under  axial  pressure  a  column  tends  to  take  the  form  shown 
in  (a),  Fig.  52.  The  hoops  prevent  this  lateral  deformation 
and  stresses  in  both  steel  and  concrete  result.  In  every  part 
of  the  section,  as  in  (c),  the  axial  force,  P,  produces  forces  in  all 
directions  in  one 
plane,  which  may 
be  resolved  into 
P'  and  P"  at  right 
angles  with  each 
other  and  in  a 
plane  perpendicu- 
lar to  the  direction 
in  which  P  acts. 
It  is  readily  seen  that  the  pair  of  forces,  P',  produces  uP', 
opposing  P",  and  so  the  force  exerted  by  the  hoops,  necessary 
to  restrain  the  deformation  of  the  concrete  laterally,  is  not 
greater  than  that  to  restrain  it  in  one  direction.  If  the  axial 
unit  stress  at  a  section  be  C,  the  lateral  unit  stress  at  a  diam- 
eter, as  shown  in  (b),  is  uC,  which  produces  a  lateral  deforma- 
tion perpendicular  to  the  diameter  equal  to  uC/Ec.  This 
deformation  is  the  same  in  the  hoops  as  in  the  concrete,  and 
the  stress  produced  is  determined  by  Es.  Then,  as  explained 
above,  this  deformation  and  stress  slightly  exceed  those  actually 
present  in  the  steel,  so 

S_        S       uC 
Es  ~  nEc       Ec 

and  S<unC 

Letting  u  =  yV,  C  =  600  Ib.  sq.  in.  and  n  =  15,  this  reduces  to 
£<9001b.  per  sq.  in. 


(a) 


FIG.  52. 


112 


REINFORCED  CONCRETE 


Experiments  confirm  the  formula  in  showing  that  the  stress 
in  the  hoops  is  very  small  for  working  loads  on  the  concrete. 
As  C  increases,  u  and  n  become  larger,  and,  in  extreme  cases, 
these  may  be  2000  Ib.  per  sq.  in.,  i  and  20  respectively.  Then 
S  will  be  about  8000  Ib.  per  sq.  in. 

Tests  of  Hooped  Columns.  Since  1905  many  series  of  tests 
on  this  type  of  column  have  been  made  and  results  published 
by  some  of  the  institutions  named  in  Chapter  III.  In  the  main 
they  each  confirm  the  others  rather  than  indicate  distinctly 
new  methods  or  results  of  investigation.  The  tests  selected 
for  illustration  here  apply,  as  many  others  do  not,  to  columns 
reinforced  only  with  hoops  or  spirals  without  longitudinal  rods. 
The  thin  strips  of  metal,  used  to  maintain  the  even  spacing  of 
the  hoops,  are  not  capable  of  taking  a  material  amount  of  com- 
pression, since  they  are  practically  on  the  outside  of  the  column. 
For  a  very  full  description  of  these  tests  and  a  thorough  dis- 
cussion of  the  results  the  reader  is  referred  to  Bulletin  No.  20, 
University  of  Illinois  Engineering  Experiment  Station,  1907. 

SUMMARY  OF  TESTS  OF  BAND-HOOPED  COLUMNS 
12  inch  diameter  columns.     Length  10  feet 


Strength  of 

Increase  in 

Reinforcement 

column 

strength  per  1% 

Ib.  per  sq.  in. 

of    reinforcement 

Initial 

Age 

coefncieDt 

Kind 

Per 
cent. 

days 

Rein- 
forced 
diame- 
ter 

Plain 
(from 
diagram) 

Lb.  per 

sq.  in. 

Per 
cent,  of 
increase 

of  elasticity 

No.  16,  2  in.,  ctoc 

1.085 

60 

2384 

1600 

725 

45 

2,670,000 

u              <«            « 

1.085 

57 

2150 

1400 

690 

49 

2,340,000 

<t                «             « 

1.050 

59 

2182 

1450 

695 

46 

2,640,000 

No.  12,  2  in.,  c  to  c 

2.081 

63 

2860 

1600 

605 

38 

2,910,000 

«          «        « 

2.071 

69 

2660 

1200 

705 

59 

2,000,000 

U                     i(                  « 

2.120 

60 

3110 

1750 

640 

37 

2,920,000 

No.  8,  2  in.,  c  to  c 

3.22 

60 

3000 

1250 

545 

44 

2,280,000 

«               «            a 

3.20 

66 

3715 

1650 

645 

39 

2,670,000 

a              «            « 

3.20 

63 

2890 

1900 

310 

16 

3,160,000 

No.  12,  3  in.,  c  to  c 

1.39 

60 

2735 

1650 

780 

47 

3,000,000 

No.  12,  4  in.,  c  to  c 

1.02 

59 

2275 

1600 

660 

41 

2,670,000 

«          «        « 

1.02 

66 

2178 

1500 

665 

44 

3,000,000 

ANALYSIS  OF  STRESSES 
9  inch  diameter  columns.     Length  10  feet 


113 


1  No.  16,  2  in.,  c  to  c 

1.35 

70 

1345 

650 

515 

79 

1,080,000 

•     u       «      «< 

1.41 

56 

1260 

600 

470 

78 

1,200,000 

«     «    « 

1.47 

65 

2140 

1200 

640 

53 

2,400,000 

«       «     « 

2.73 

67 

2970 

1600 

500 

31 

2,670,000 

2  No.  12,  2  in.,  c  to  c 

2.94 

67 

3561 

1950 

550 

28 

3,550,000 

2     «       «      « 

2.80 

81 

3685 

2150 

550 

26 

3,910,000 

SUMMARY  OF  TESTS  OF  SPIRAL-HOOPED  COLUMNS 
Length,  10  feet.     Diameter,  12  inches.     Pitch,  1  inch 
High  Carbon  Steel 


Increase  in  strength 

Reinforcement 

Strength  of  column 

per  1%  of  rein- 

forcement 

Age 

Initial 

CO6ffici6Dt  OI 

Size  of  wire 

Per 
cent. 

days 

Reinforced 

Plain 
(from 
diagram) 

Lb.  per 

sq.  in. 

Per  cent 
of 
increase 

elasticity 

No.  7 

0.85 

56 

2503 

1600 

1060 

66 

2,670,000 

u 

0.85 

63 

2506 

1700 

950 

56 

2,840,000 

3       « 

0.82 

57 

2010 

1400 

— 

— 

2,800,000 

£inch 

1.73 

57 

2718 

1400 

760 

54 

2,000,000 

« 

1.67 

60 

3800 

1900 

1140 

60 

2,720,000 

u 

1.68 

63 

3793 

1900 

1130 

59 

2,540,000 

Mild  Steel 

No.  7 

0.84 

60 

2080 

1350 

870 

64 

2,250,000 

u 

0.85 

64 

2203 

1500 

830 

55 

2,500,000 

it 

0.84 

61 

2220 

1450 

920 

63 

2,420,000 

|inch 

1.64 

58 

2068 

1150 

560 

49 

1,920,000 

a 

1.71 

62 

3404 

1800 

940 

52 

3,280,000 

M 

59 

~ 

~ 

2 

— 

A  1-2-4  mixture  was  used  in  every  case. 

1 1-4-8  mixture.        2 1-1^-3  mixture.     All  others  were  1-2-4  mixture. 

3  This  column  was  loaded  with  2000  pounds  per  sq.  in.,  when  the  hoops 
were  stripped  and  it  was  tested  as  a  plain  column  with  the  result  given  in  the 
table. 


114 


REINFORCED  CONCRETE 


The  seventh  column  of  the  table  indicates  per  cent,  of  increase 
in  strength  of  the  reinforced  over  the  plain  concrete  (as  deter- 
mined from  the  diagram  below)  for  each  1  per  cent,  of  reinforce- 
ment, and  is  found  by  dividing  column  six  by  column  five. 

The  strength  of  a  plain  column  of  the  same  material  is  approxi- 
mately found  from  the  stress-strain  diagram  plotted  for  each 
test  of  a  reinforced  column.  The  diagrams  for  two  of  the  speci- 
mens are  shown  here.  Fig.  53  being  for  the  sixth  in  the  Table 
for  Band-Hooped,  and  Fig.  54  for  the  fifth  in  the  last  table 
above.  As  is  common  to  all  such  diagrams,  there  is,  within 
narrow  limits,  a  decided  change  in  the  slope  of  this  curve,  the 
first  part  representing  the  action  of  the  column  before  the  hoops 
are  brought  into  tension.  The  next  stage  is  represented  very 


=--d 


^ 

y& 

^-*- 

^^ 

—   — 

2800 

^ 

*^ 

*  __^^ 

^ 

2000 

">" 

/ 

\  / 

1200 

\/ 

V 

400 

Delo 

rmat^on  p 

er  unit  of(Jength 

0  0  0005  0  OJ1 

FIG.  53. 

nearly  by  a  straight  line  and  continues  until  nearly  the  ultimate 
load  is  imposed.  By  producing  this  line  backward  to  A,  on 
an  ordinate  through  the  origin,  the  load  is  found  that  is  very 
nearly  the  ultimate  for  the  plain  column  as  determined  by  other 
tests  on  such  columns. 

An  inspection  of  these  or  similar  stress-strain  diagrams  shows 
nearly  a  straight  line  until  the  unit  load  imposed  is  consider- 
ably in  excess  of  the  working  stresses;  next,  a  rapid  change 
in  inclination  of  the  curve  until  the  unit  deformation  is,  perhaps, 
0.0015,  during  which  time  a  readjustment  of  stresses  takes 
place  between  the  concrete  and  the  steel;  then,  a  period  during 
which  stresses  and  deformations  change  in  like  ratio,  and  finally 
a  period  of  ultimate  failure  with  large  deformations  and  small 
increase  or  some  decrease  in  the  loads. 

The  ultimate  axial  deformations  are  influenced  somewhat  by 
the  fact  that  there  is  considerable  lateral  deformation  in  the 


ANALYSIS  OF  STRESSES 


115 


column,  especially  as  the  ultimate  load  is  approached.  Figs.  49, 
50,  and  51  show  the  appearance  of  specimens  of  a  band-hooped 
and  likewise  of  a  spiral-hooped  column  failure.  The  very 
decided  change  in  the  value  of  the  coefficient  of  elasticity 
from  the  initial  to  the  ultimate,  of  course,  is  a  consequence  of 
the  conditions  stated.  The  stress-strain  diagrams  for  all  the 
tests  recorded  in  the  tables  cannot  be  reproduced  here,  but 
it  may  be  said  that  the  shortening  of  the  column  under  the 
load  depends  upon  the  amount  and  type  of  reinforcement, 
being  more  with  spirals  than  with  hoops.  As  compared  with 
the  shortening  of  a  plain  column,  the  spiral,  for  each  one 
per  cent,  of  reinforcement  allows  about  twice  the  excess  of 
axial  deformation,  as  does  the  band-hoops,  but,  at  the  same 


I)elormatipn  per  unit  ot  length 


2000 


0.0005 

FIG.  54. 


0.001 


time,  will  sustain  materially  greater  loads,  as   shown  by  the 
tables. 

If  the  coefficients  of  elasticity  of  the  hooped  columns  be  com- 
pared with  those  for  the  plain  concrete  columns  in  the  table 
on  page  106,  the  rather  remarkable  fact  is  apparent  that  the 
hooping  does  not  serve  to  diminish  deformation  in  a  column 
even  under  working  loads.  The  average  value  of  Ec  from  the 
table  for  plain  columns  is  3,090,000  Ib.  per  sq.  in.  for  the 
1:2:4  mixture,  while  for  banded  reinforcement  it  is  2,666,000, 
and  for  the  spiral  2,540,000  Ib.  per  sq.  in.  There  seems  to  be 
no  known  reason  for  this  action  of  the  reinforcement,  unless  it 
be  that,  during  fabrication,  more  perfect  compacting  of  the 
plain  concrete  is  possible  without  it.  However,  the  shortening 
of  a  column,  of  such  length  as  is  found  in  engineering  practice, 
is  not  materially  different  whether  Ec  be  1,500,000  or  3,000,000 


110  REINFORCED  CONCRETE 

Ib.  sq.  in.  For  example,  let  the  unit  stress  be  600  Ib.  sq.  in. 
and  the  length  25  feet.  Then  the  shortening  will  be  300  X  600 
-=-  1,500,000  or  300  X  600  -f-  3,000,000  =  0.12  inch  or  0.06 
inch,  according  to  the  value  of  Ec  used. 

During  the  above  tests  the  hooped  columns  shortened  from 
|  inch  to  If  inches  in  the  10  feet  of  length,  while  the  lateral 
deflection  was  apparent  even  before  the  maximum  loads  were 
imposed.  In  some  cases  this  lateral  deformation  was  as  much 
as  5  inches,  after  the  period  of  maximum  load  had  been  passed, 
but  while  the  column  was  in  one  piece  and  still  capable  of  bear- 
ing some  load.  Columns  of  this  kind  are  actually  broken  into 
two  pieces  with  considerable  difficulty. 

The  Strength  Added  by  Hoops.  Referring  to  the  above  tables 
of  tests,  it  is  seen  that  the  gain  in  strength  of  the  12-inch 
banded  columns  over  the  plain  concrete  is  639,  or,  omitting  the 
310  as  belonging  to  a  column  known  to  be  exceptional,  669 
Ib.  sq.  in.  for  each  one  per  cent,  of  reinforcement.  The  corre- 
sponding increase  for  the  9-inch  columns  is  538  Ib.  per  sq.  in. 
The  next  column  of  figures  shows  the  per  cent,  of  increase  for 
each  one  per  cent,  of  reinforcement.  The  change  of  mixture 
to  1-4-8  and  1-1 J-3  seems  to  have  but  little  effect  upon  the 
strength  added  to  the  9-inch  columns  by  the  steel,  but  is  shown 
distinctly  in  "per  cent,  of  increase." 

The  spiral  hooping  adds  more  than  do  the  bands  to  the  col- 
umn strength,  the  increase  being  an  average  of  1008  and  890 
Ib.  sq.  in.  for  the  high  carbon  and  mild  steel  respectively,  the 
next  to  the  last  test  being  omitted  as  it  was  known 'to  be  not 
representative.  With  this  large  additional  strength  was  a 
greater  ultimate  compression,  more  lateral  deflection  of  the  axis 
of  the  column,  and  perhaps  more  squeezing  out  of  the  concrete 
between  the  hoops. 

The  author  of  the  Bulletin  concludes  that  the  general  formula, 
given  on  page  103  may  be  adapted  to  use  by  substituting  the 
following  constants.  Then  for  band-hooped  columns, 

C'  =  1600  +  65,000  p  (106) 

and  for  spiral-hooped  columns 

Cf  =  1600  +  100,000  p  (107) 

giving  about  an  average  of  results  for  mild  and  high  carbon 


ANALYSIS  OF  STRESSES  117 

steel.  These  formulas  are  only  to  indicate  the  ultimate  and 
not  the  working  loads. 

Working  Loads  for  Hooped  Columns.  The  Joint  Committee 
(see  pages  151  and  152)  recommend  that,  when  the  length  does 
not  exceed  12  diameters,  the  working  stress  may  be  450  X  1.20 
=  540  Ib.  per  sq.  in.  if  the  plain  concrete  be  able  to  hold  2000 
Ib.  per  sq.  in.,  and,  in  general,  the  working  stress  may  be  taken 
as  1.20  X  22.50  per  cent,  of  the  ultimate  strength  of  the  con- 
crete at  the  age  of  28  days.  It  will  be  noted  that  this  is  not 
far  from  what  would  result  if  the  ultimate  strength  by  the  above 
formulas  be  divided  by  four. 

Columns  with  Hooped  and  Longitudinal  Reinforcement.  The 
formula  of  Considere,  given  above,  page  (87),  is  intended  to 
apply  to  this  type  of  construction,  S'A2  being  the  expression 
for  the  strength  added  by  the  rods.  To  make  his  formula  con- 
form to  the  instructions  issued  by  the  Minister  of  Public  Works 
of  France,  Considere,  in  1906,  published  a  modified  form  of 
his  expression,  which  is,  in  pounds  per  square  inch, 

d  =  C  +  34,000  (p  +  2.1  p'}  (108) 

in  which  p  and  p'  are  proportions  of  steel  found  as  explained 
above,  C\  is  the  unit  strength  of  the  reinforced,  and  C  that  for 
the  plain  column.  If  the  section  be  octagonal  and  the  steel 
not  very  near  the  outside,  the  strength  of  the  area  of  concrete 
inside  the  steel  may  be  multiplied  by  1.5  for  the  total  strength 
of  the  concrete. 

Such  formulas  have  particular  reference  to  the  ultimate  rather 
than  the  working  load,  although  the  effect  of  the  rods  is  felt 
from  the  first  application  of  stress.  To  get  the  full  effect  of 
rods  the  joints,  if  any,  should  be  carefully  made,  either  by 
lapping  them  so  as  to  secure  the  proper  bond  or  by  butting 
them  closely  together  at  the  ends.  If  the  rods  be  large  and 
correspondingly  few  in  number  it  is  sometimes  advisable  to 
provide  sleeves  to  secure  the  continuity  of  the  steel.  If  the 
rods  be  placed,  at  least,  1J  inches  inside  the  surface,  they  sel- 
dom buckle  under  working  loads,  and  the  hooping,  no  doubt, 
serves  to  prevent  any  such  tendency.  For  this  purpose  the 
amount  of  steel  in  the  bands  seldom  needs  be  as  much  as  one 
per  cent. 

It  seems  probable  that  the  best  results  are  obtained  when 


118 


REINFORCED  CONCRETE 
CONSIDERED  COLUMNS 


No. 

Spiral 
reinforcement 

Longitudinal 
reinforcement 

Per 
cent 
spiral 
rein- 
force- 
ment 

Per 
cent 
longi- 
tudinal 
rein- 
force- 
ment 

Per 

cent 
total 
rein- 
force- 
ment 

Unit 
stress 
at  first 
crack 
Ib.  per 
sq.  in. 

Ulti- 
mate 
unit 
stress 
Ib.  per 
sq.  in. 

Ratio 
strength 
of  rein- 
forced to 
plain 
con- 
crete 

Pitch 
inches 

Diam- 
eter 
inches 

No. 

rods 

Diam- 
eter 
inches 

1 

None 

None 

None 

None 

None 

None 

1892 

1892 

1.0 

2 

1.50 

T36 

4 

i 

4 

0.63 

0.25 

0.88 

2262 

2262 

1.19 

3 

1.48 

1- 

4 

\ 

1.25 

0.25 

1.50 

2290 

2532 

1.34 

4 

1.65 

i 

4 

i 

2.20 

0.25 

2.45 

2418 

3414 

1.80 

5 

1.50 

A 

8 

& 

0.63 

1.22 

1.85 

3186 

3215 

1.70 

6 

1.46 

i 

8 

A 

1.25 

1.22 

2.47 

3272 

3271 

1.73 

7 

1.69 

1 

8 

A 

2.20 

1.22 

3.42 

3442 

3997 

2.12 

8 

1.22 

\ 

4 

i 

4 

1.49 

0.25 

1.74 

2788 

2845 

1.50 

9 

1.58 

I 

4 

1 

2.32 

0.25 

2.57 

2418 

3001 

1.59 

10 

1.61 

i 

4 

1 

4 

3.30 

0.25 

3.55 

2560 

3641 

1.92 

11 

1.46 

A 

4 

i 

5.00 

0.25 

5.25 

2247 

3499 

1.85 

12^ 

1.58 

\ 

8 

• 

Tff 

1.16 

0.25 

1.41 

2318 

2318 

1.22 

122 

1.58 

I 

8 

1 

2.32 

0.49 

2.81 

2333 

3271 

1.73 

123 

1.58 

i9s 

8 

1 

4.62 

1.01 

5.63 

2617 

4295 

2.27 

13* 

3.15 

\ 

8 

i 

0.58 

0.49 

1.07 

2304 

2304 

1.22 

132 

3.15 

I 

8 

1 

1.18 

1.01 

2.19 

2546 

2574 

1.36 

13s 

3.15 

A 

8 

i 

2.31 

1.45 

3.76 

2646 

2830 

1.50 

141 

4.72 

i 

8 

I 

0.39 

1.01 

1.40 

2205 

2205 

1.16 

142 

4.72 

f 

8 

i 

0.78 

1.45 

2.23 

2603 

2603 

1.37 

143 

4.72 

A 

8 

A 

1.54 

1.96 

3.50 

2944 

2944 

1.56 

the  reinforcement  is  about  equally  divided  between  the  hooping 
and  the  longitudinal  rods. 

Tests  of  Columns  with  both  Longitudinal  and  Hooped  Rein- 
forcement. The  following  tests,  referred  to  before  on  page 
109,  were  made  on  octagonal  columns  of  625  cm2  section  and 
one  meter  long.  They  were  conducted  for 
the  owners  of  the  German  rights  of  Consi- 
dere's  patents  at  a  building  site,  and  without 
effort  to  exceed  the  care  usually  attained  in 
construction.  Fig.  55  shows  a  cross-section 
of  one  of  these  columns.  The  tests  were 
made  under  the  direction  of  Professor  Bach, 
FIG.  55.  of  The  Royal  Technical  High  School  of  Stutt- 


ANALYSIS  OF  STRESSES 


119 


gart,  and  the  results  were  published  in  1905.  The  fact  that 
the  length  is  short  in  comparison  with  the  diameter  makes 
the  results  of  the  tests  of  less  practical  value  than  would  have 
been  the  case  with  longer  samples. 

The  diameters  of  the  wire  are  given  in  the  nearest  English 
equivalent  of  the  metric  dimensions  and  so  are  not  exact.  The 
percentage  of  steel  is  computed  on  the  basis  of  the  whole  area 
of  the  cross-section  as  the  hoops  were  1 J  inches  inside  the  sur- 
face. By  comparing  the  added  strength  with  the  computed 
addition  according  to  formula  (108)  it  is  seen  that  there  is  a 
fair  agreement  while  the  pitch  remains  about  y,  but  the  measured 
strength  does  not  equal  the  computed  in  the  last  six  tests.  The 
spiral  does  not  seem  to  be  2.4  times  as  effective  as  the  longi- 
tudinal reinforcement  when  the  percentage  of  the  former  is 
large  as  compared  with  the  latter. 

It  is  not  possible  to  determine  just  what  is  the  effect  of  the 
hoops  or  of  the  rods,  but  it  seems  that  the  latter  may  be  effect- 
ively used  to  the  same  extent  as  the  spirals  as  is  shown  by  Nos. 
5,  6,  and  7. 

The  following  tests  were  made  at  the  Watertown  Arsenal  in 
1905,  and  were  published  in  " Tests  of  Metals"  in  1906.  The 
hoops  were  riveted,  the  concrete  was  1:2:4  mixture,  the  diam- 
eter was  in  each  case  10|  inches,  and  the  length  was  8  feet. 
The  columns  were  5  to  6  months  old  when  the  tests  were  made. 
The  longitudinal  reinforcement  was  made  up  of  1  X  1  X  | 
inch  angles;  the  hoops  were  0.12  X  1.5  inches. 

COLUMNS  WITH  HOOPS  AND  LONGITUDINAL  ANGLES 


Reinforcement 

Per  cent, 
hoops 

Per   cent, 
angles 

Strength 
Ibs.  per 
sq.  in. 

Increase 
over    plain 
column 

Increase 
over 
hooped 
column 

Per  cent,  of  in- 
crease per 
1  per  cent,  of 

Number 
hoops 

Number 
angles 

Hoops 

Angles 

None 

None 

None 

None 

1413 

13 

0 

1 

0 

2232 

819 

— 

58 

— 

13 

4 

1 

1.0 

3029 

1616 

797 

58 

56 

25 

0 

1.8 

0 

3428 

2015 

— 

79 

— 

25 

4 

1.8 

1.0 

4189 

2776 

761 

79 

54 

47 

0 

3.4 

0 

5289 

3856 

~ 

80 

It  may  be  said  that  the  longitudinal  reinforcement  is  effective 
in  the  early  stages  of  the  tests,  or  within  the  working  loads, 


120  REINFORCED  CONCRETE 

while  the  hoops  increase  the  ultimate  strength.  In  the  table 
the  beneficial  effect  of  decreasing  the  pitch  of  the  hooping  is 
readily  seen.  While  one  per  cent,  of  hooping  adds  58  per  cent. 
to  the  strength,  3.4  per.  cent,  adds  272  per  cent.,  and  the  latter 
is  38  per  cent,  more  effective  per  unit  of  cross-section.  In 
making  these  comparisons  it  should  be  noted  that  the  strength 
of  the  plain  concrete  column  was  below  normal. 

Working  Loads.  If  the  reinforcing  bars  be  firmly  imbedded 
in  the  concrete  the  elastic  limit  of  the  steel  is  a  controlling  factor 
in  determining  the  safe  strength  of  the  column.  If  the  stress 
in  the  metal  be  taken  as  J  the  elastic  limit,  or  about  8000  Ib. 
per  sq.  in.,  the  stress  in  the  concrete  will  be  this  amount  divided 
by  n,  or  530  Ib.  per  sq.  in.,  and  this  may  safely  be  used  if  p 
be  as  much  as  one  per  cent. 

The  Joint  Committee  recommends  (see  page  152)  38  per  cent. 
of  the  ultimate  strength  of  the  plain  column,  or  600  Ib.  per 
sq.  in.  for  good  grade  of  concrete. 

Long  Columns.  When  the  length  is  not  more  than  about 
eight  or  ten  times  the  diameter,  the  strength  of  the  column 
differs  but  little  from  the  cube  or  the  short  cylinder,  as  is  apparent 
from  the  records  of  tests  given  before.  Between  the  limits, 
length  =  10  diameters  and  length  =  15  diameters,  the  compu- 
tations and  recommended  unit  stresses  may  he  used  as  given 
in  this  chapter.  Beyond  the  last-named  ratio  the  action  of  the 
load  is  not  so  nearly  that  of  direct  crushing  and  the  effect  of 
the  bending  moment  must  be  recognized  from  the  first.  Con- 
crete, whether  plain  or  reinforced,  is  not  specially  adapted  to 
this  type  of  construction,  and  is  not  so  used  to  a  great  extent. 

The  ordinary  formulas  for  columns  of  other  materials  may 
theoretically  be  adapted  to  those  of  concrete  when  the  constants 
have  been  well  determined.  This  question  has  not  been  given 
the  attention  necessary  to  a  final  statement  as  to  the  value  of 
<£  in  the  formula,  but  the  form  of  the  equation  should  be 

P  C 


in  which  <£  is  the  constant  to  be  determined  for  various  condi- 
tions of  reinforcement,  mixture,  and  methods  used  in  fixing  the 
ends,  and  r  is  the  radius  of  gyration.  Some  tests  on  long 


ANALYSIS  OF  STRESSES  121 

columns  of  the  "Gray"  section  have  been  made  at  the  Uni- 
versity of  Illinois  Engineering  Experiment  Station  which  show 
that  a  straight  line  formula  may  be  used.  The  columns  were 
made  of  angles  and  tie  plates  wound  with  a  wire  spiral.  The 
steel  was  13  sq.  in.  in  cross-section  or  10.8  per  cent,  of  the 
gross  column  area.  The  columns  were  from  4  ft.  8  in.  to  19  ft. 
4  in.  in  length  and  16  inches  across  the  section,  which  included 
a  fire-proofing  covering  2  inches  thick.  The  compressive 
strength  of  cubes  of  the  concrete  was  2150  Ib.  per  sq.  in.  The 
ultimate  strength  of  the  steel  column  was 

35,000 
P  =  A 


12,000r2 
and  for  the  longest  column  Z2/r2  was  59.5. 

The  ultimate  strength  of  the  reinforced  column  was  given  by 
P  =  A  (5100  +  45  l/d) 

The  area  of  the  concrete  case  was  120  sq.  in. 

Many  of  the  tests  on  long  columns  seem  to  indicate  that  the 
loss  of  strength  with  the  increase  of  length  is  largely  due  to  the 
fact  that  the  number  of  weak  spots  in  the  column  is  propor- 
tionally increased.  Long  columns  often  break  near  or  at  the 
end  rather  than  near  the  middle. 

PROBLEMS 

50.  Find  the  size  of  a  square  column  to  have  2  per  cent,  of  longitudinal 
reinforcement  only,  and  to  sustain  a  load  of  100,000  pounds.     Assume 
1:2:4  mixture  and  mild  steel.' 

51.  The  diameter  of  a  column  is  8  inches,  and  the  imposed  load  is 
to  be  105,000  pounds.     What  is  the  required  amount  of  reinforcement 
in  the  form  of  longitudinal  bars? 

52.  A  square  column  10  by  10  inches  carries  a  load  of  125,000.     If 
the  reinforcement  be  4  f -inch,  round  rods,  what  is  the  stress  in  the  steel 
and  in  the  concrete? 

53.  What  should  be  the  diameter  of  a  column,  reinforced  with  1  per 
cent,  of  hooping,  safely  to  carry  a  load  of  90,000  pounds? 

54.  The  load  on  a  column  is  to  be  200,000  pounds.     If  the  diameter 
be  18  inches,  what  per  cent,  of  hooped  reinforcement  is  required  to  keep 
the  stresses  within  safe  limits? 

55.  If  the  load  be  75,000  pounds,  the  diameter  7  inches,  p  =  1  per 
cent.,  find  the  stresses  in  the  concrete  and  in  the  steel. 


122  REINFORCED  CONCRETE 

56.  Required  the  area  of  a  column  10  feet  long  carrying  a  load  of 
400  tons,  having  3  per  cent,  of  longitudinal  and  1  per  cent,  of  hooped 
reinforcement. 

57.  In  Problem  51  what  amount  of  reinforcement  will  be  sufficient 
if  it  be  equally  divided  between  hoops  and  longitudinals? 

Beams  Supported  in  More  than  Two  Places.  Under  this  head 
may  be  included  continuous  beams,  or  those  supported  at  various 
points  in  the  span,  and  also  those  supported  along  lines  not 
parallel  to  each  other,  as  slabs  resting  on  three  or  more  edges. 

Continuous  Beams.  When  the  bending  moments  at  all  points 
in  a  continuous  beam  are  known,  the  beam  is  designed  or  inves- 
tigated by  the  same  formulas  as  are  used  for  the  purpose  with 
simple  beams.  These  moments,  in  turn,  depend  upon  the 
moments  at  the  various  supports  which,  for  uniform  loads,  are 
determined  by  the  application  of  what  is  called  "the  theorem 
of  three  moments,"  as  deduced  by  Clapeyron  in  1857.  This 
theorem  *  assumes  that  the  supports  are  all  on  the  same  level, 
and  that  the  moments  of  inertia  of  the  cross-sections  are  uni- 
form throughout  all  the  spans.  In  reinforced  concrete  con- 
struction the  first  assumption  is  usually  correct,  the  second  one 
often  is  not.  The  moments  at  the  supports  due  to  concentrated 
loads  may  also  be  found  by  method  of  reasoning  not  very  different 
from  that  for  uniform  loads,  and  it  is  not  necessary  to  present 
either  method  here  in  great  detail. 

By  means  of  the  theorem  of  three  moments  a  general  formula 
for  the  negative  bending  moments  at  any  support,  for  equal 
spans  and  uniform  loads,  may  be  deduced.2  This  formula  is 

Mr  =  -  *r-*D-r+i-Dr-,L.-r  wP  (109) 

2  A»~l 

in  which  r  is  the  number  of  the  support  from  the  left,  and  n 
is  the  number  of  spans.     Also 

A0  =  1,  A!  =  4,  D0  =  0  and  D,  =  1. 
In  general  Ar  =  4  \_l  —  Ar_2 

and  Dr  =  Ar-1  -  Dr_,. 

1  See  Merriman's  Mechanics  of  Materials,  Chap.  VIII. 

2  See  article  by  A.  R.  Crathorne  in  Science,  April  29,  1910. 


ANALYSIS  OF  STRESSES  123 

Making  these  substitutions  the  following  values  of  A  and  D 
are  found. 

A0  =  l  A3  =  56  D0  =  0  D,  =  12 

A4  =  209  Di  =  1  D4  =  44 

A5  =  780  D2  =  3  '  Z)5  =  165 

Other  values  may  be  easily  added  if  required. 

As  an  example,  let  it  be  required  to  find  the  bending  moment 
at  the  third  support  from  the  left  in  a  continuous  beam  of  five 
spans.  Here  M3  =  -  (AiD3  -  DA)  -f-  2  A4  wp  =  3/38  wl2. 

The  positive  moment  at  x  to  the  right  of  a  support  is  M'  — 
V'x  +  ^  wx2  where  M'  is  the  moment  at  the  support  to  the  left 
of  the  section  and  V  is  the  shear  just  to  the  right  of  the 
same  support.  The  value  of  V  is  found  from 

VI  =  M"  -  M'  +  \  wl2 

in  which  M"  is  the  moment  at  the  support  to  the  right  of  the 
given  section.  The  shear  at  the  other  end  of  the  panel  is  always 
wl  -  V. 

The  maximum  positive  moment  is  at  x  =  V  -f-  w,  and  becomes 

M  =  M'n  +  7'n/2  w 

The  subscript  indicates  the  number  of  the  panel  considered. 

Continuous  beams  of  reinforced  concrete  are  nearly  always 
built  with  fixed  ends  and  are  rigidly  attached  to  the  intermediate 
supports.  That  the  latter  fact  is  not  considered  in  the  com- 
putations of  the  moments  is  a  practice  on  the  side  of  safety. 
If  the  beam  of  four  spans  be  fixed  at  the  ends  and  supported 
in  uniform  spans,  with  a  uniform  load  on  all  the  spans,  the 
maximum  negative  bending  moment  at  the  supports  is 

M  =  -  TV  wl* 
and  the  maximum  positive  moments  between  supports  is 

M  =  A  wl2. 
If  the  ends  be  simply  supported,  the  maximum  moments  become 

M  =  -  A  wl2. 
and  M  = 


124  REINFORCED  CONCRETE 

For  either  more  or  fewer  spans,  generally,  the  moments  will 
be  less  than  these.  It  is  possible,  by  omitting  the  loads  on  a 
part  of  some  spans,  to  slightly  increase  these  moments,  but 
this  fact  is  partly,  at  least,  balanced  by  the  rigidity  of  the  con- 
nections between  beam  and  the  columns  or  other  beams  forming 
the  supports.  In  practice,  moreover,  it  is  impossible  to  omit 
all  of  the  load  on  any  panel  as  the  weight  of  the  concrete  itself 
is  considerable. 

Working  Moments  for  Continuous  Beams.  In  view  of  the 
above-named  considerations  it  has  become  the  practice  to  make 
the  maximum  moments,  both  positive  and  negative,  TV  wl2, 
except  at  the  ends  of  the  beam  and  at  the  middle  of  the  end 
panels  where  the  moments  are  taken  as,  yV  wl2.  In  these  expres- 
sions w  is  both  the  live  and  the  dead  load  per  linear  foot.  If 
the  beam  be  a  slab  continuous  over  several  floor  beams  the 
moments  are  uniformly  yV  wZ2.  These  figures  are  in  almost 
universal  use,  both  in  this  country  and  in  Europe,  and  are  be- 
lieved to  represent  conservative  practice  if  the  concrete  be 
well  made  and  if  both  that  and  the  steel  be  erected  in  a  work- 
manlike manner. 

Slabs  Supported  on  Four  Sides.  This  question  involves  two 
problems,  (a)  the  stresses  in  the  slab  itself,  and  (6)  the  distribu- 
tion of  the  loads  carried  by  the  slab  upon  the  supporting  beams. 
Reliable  tests  and  thorough  analyses  have  been  made  concerning 
the  stresses  in  slabs  of  homogeneous  material  by  Bach,  Grashof, 
and  others,  but  they  are  not,  in  all  respects,  applicable  to  rein- 
forced concrete.  There  are  numerous  difficulties  in  the  way 
of  making  many  tests  of  full-sized  floor  slabs  of,  say,  15  feet 
square.  As  a  result  few  such  tests  have  been  made  in  such  a 
way  as  to  be  useful  as  references. 

The  usual  method  employed  in  computing  the  stresses  in  a 
slab,  supported  or  fixed  around  the  edges,  and  reinforced  with 
rods  parallel  to  the  sides,  is  to,  first,  find  the  proportion  of  the 
whole  load  carried  by  each  system  of  reinforcement,  and  then 
consider  the  slab  a  simple  beam  or  fixed  beam,  as  the  case  may 
be.  This  method  is  only  approximate,  but  stresses  so  computed 
should  be  larger  than  those  actually  existing. 

Let  Fig.  56  represent  a  slab,  of  length  I  and  breadth  6,  resting 
upon  beams  on  the  four  sides.  Let  the  reinforcement  consist 
of  rods  parallel  to  I  and  to  6,  and  let  the  load  be  uniform  over 


ANALYSIS  OF  STRESSES 


125 


the  area  bL  It  is  required  to  find  Wi  and  w2,  the  parts  of  the 
whole  unit  load,  w,  that  is  carried  by  the  reinforcement  parallel 
to  xx  and  by  that  parallel  to  yy.  The  deflections  of  the  strips 
of  unit  width  at  a,  a  point  on  a  diagonal,  are,  of  course,  the  same 
for  xx  as  for  yy,  and  are  proportional  to  the  fourth  power  of 
the  length  of  the  strip.  Hence,  Wi64  =  w2/4,  and,  since  Wi-\-w2  —  w 


or 


w  — 


w 


which  represents  the  proportion  of  the  load  carried  by  the 
system  parallel  to  the  shorter  axis.  From  this  formula  the 
following  table  is  easily  deduced, 


I 

1.0 

1.1 

1.2 

1.3 

1.4 

1.5 

2.0 

b 

IVi 

.50 

.59 

.67 

.74 

.80 

.84 

.89 

w 

1 


i\ 

x                         ^ 

y             a 

- 

X 

FIG.  56. 


Thus,  if  a  slab  be  10  X  12,  the  system  parallel  to  the  shorter 
axis  carries  more  than  two- 
thirds  of  the  load.  The  steel 
parallel  to  the  longer  axis  will 
not  be  stressed  sufficiently  to 
make  its  use  economical  if  the 
disproportion  in  the  lengths 
of  the  sides  be  great.  For 
this  reason  it  is  common  to 
use  the  longer  system  only  to 
provide  for  shrinkage  and  expansion  stresses  if  l/b  be  more 
than  1.5,  and  to  provide  for  the  carrying  of  the  entire  load 
on  the  transverse  system. 

Under  a  uniform  load  the  slab  deflects  from  a  plane  to  a  curved 
surface.  The  intersection  of  this  surface  and  a  plane  perpen- 
dicular to  the  slab  is  a  curve,  the  equation  of  which  depends 
upon  the  amount  and  distribution  of  the  reinforcement.  Were 
this  equation  known  it  would  be  possible  to  compute  the  stress 
at  any  point  from  the  deformation,  and  the  reinforcement  could 
be  distributed  as  needed.  It  is  readily  seen  that,  in  the  beam 


126  REINFORCED  CONCRETE 

of  length  6  and  width  I,  the  supports  parallel  to  b  diminish  the 
stresses  along  those  sides  and,  hence,  less  reinforcement  is  needed 
in  the  stripps  xx,  that  are  near  the  ends  than  in  those  near  the 
middle.  Even  without  exact  computation  some  difference  in 
the  spacing  of  the  rods  may  be  made,  and  the  saving  of  steel 
thus  possible  is  considerable,  but  the  spacing  is  usually  even. 

Having  determined,  from  the  table,  the  proportion  of  the 
load  carried  by  the  transverse  reinforcement,  the  moment  is 
J  wib2  if .  the  slab  be  simply  supported.  If  the  slab  be  con- 
tinuous the  moment  will  be  yV  wib2  or  iV  w i62,  according  to  the 
rules  laid  down  under  continuous  beams  above.  The  recom- 
mendations of  the  Joint  Committee  will  be  found  on  page  148, 
and  are  to  be  followed  in  design  and  investigation.  It  will 
be  found  that  panels  should  be  either  comparatively  long  and 
narrow,  so  that  one  system  of  reinforcement  can  take  all  the 
load,  or  else  square,  so  that  the  load  may  be  evenly  divided  be- 
tween the  two  systems  and  the  working  strength  of  steel  may 
be  fully  developed  in  both. 

Distribution  of  Floor  Loads  to  Beams.  If  but  one  system  of 
reinforcement  be  used,  all  the  load  on  the  slab  is  considered  to 
be  carried  to  the  beams  at  the  ends,  and  the  uncertain  part 
that  is  carried  to  the  girders  supporting  the  beams  is  ignored 
in  the  computations.  The  same  assumption  will  be  made  in 
case  the  slab  is  reinforced  in  two  directions  unless  it  be  nearly 
square.  In  such  cases  it  is  certain  that  this  assumption  will 
not  hold  good. 

Fig.  57  represents  a  slab,  nearly  square,  resting  upon  the 
four  surrounding  beams.  If  the  slab  be  reinforced  in  both 
directions  the  load  will  be  carried  to  the  two  sets  of  parallel 
beams  as  indicated  in  the  above  table,  page  125.  Of  the  part 
of  the  load  that  is  carried  to  the  beams  ad  and  6c,  one-half  is, 
evidently,  supported  by  each  of  these  beams  if  the  load  be 
uniform.  If  the  beams  ab  and  dc,  with  the  part  of  the  load 
carried  by  them,  are  to  be  considered  quite  separately  from 
the  other  two,  the  load  on  beam  ad  is  represented  by  adfe,  uni- 
formly distributed  along  ad.  The  effect,  however,  of  the 
support  afforded  by  beams  ab  and  dc  is  to  prevent  a  uniform 
distribution  of  a  uniform  load.  If  the  relations  between  the 
deflection,  the  bending  moment,  and  the  reactions  of  a  beam 
with  a  uniform  load  be  considered,  it  is  at  once  apparent  that 


ANALYSIS  OF  STRESSES 


127 


the  loads  are  not  uniform  along  the  beams.  Let  a  section  be 
passed  through  ef,  midway  between  ad  and  cb,  cutting  the  loaded 
slab  in  some  unknown  curve.  The  longest  ordinate  to  this 
curve  will  be  at  the  middle.  Let  narrow  strips,  xx,  be  passed, 
intersecting  ef  at  various  points,  where  the  deflections  on  xx 
will  be  the  same  as  on  ef.  The  strip,  xx,  passing  through  the 
middle  of  ef,  has  a  greater  deflection  and,  hence,  a  greater  reaction 
on  ad  than  have  strips  nearer  ab  and  dc.  If  the  curve  of  the 
loaded  slab  surface  were  known,  the  reaction  of  the  strips  could 
be  determined  with  consider-  §  ,  (  ,  ( 

able  certainty,  and  even  with- 
out such  knowledge  it  is  clear 
that  the  loads  are  greater 
toward  the  middle  of  the 
beams. 

Fig.  57  shows  three  assump- 
tions as  to  the  distribution  of 
loads    from    the    slags.     The 
curve  ahd  is  a  parabola,  and 
agd   a   triangle,   each  having 
the   area   aefd,   which   repre-    — , 
sents   the   weight    carried  to       ' 
the  beam  ad,  if  the  thickness 


b                   ii      A» 

V                                     M         /            \ 

1  1  /    \ 
1  1/       \ 

yl                \ 

'7  11            ^\ 

c 

-p            i~f-    -\r                      -\  v 

/          \\ 

//          X 

//           > 

'/                1 

i/a                  1  1 

f 

i 

\ 

£j 

i  r 


FIG.  57. 


of  the  slab  and  the  load  be  uniform.     The  rectangle  aefd  repre- 
sents a  uniform  distribution  of  the  load. 

If  the  load  be  applied  as  ordinates  of  the  parabola  the  bending 
moment  at  the  middle  of  the  longer  beam  will  be 


M  = 


wl*     bl 

4  V2 


64  + 


or  for  the  shorter  beam 


M  = 


(HO) 


(HI) 


in  which  w  is  the  uniform  load  per  unit  of  area  on  the  slab,  and 
Wi  and  wz  are  the  parts  of  that  unit  load  that  go  to  the  longer 
and  shorter  beams  respectively,  as  given  in  the  table  on  page 
125.     If  the  slab  be  square,  Wi  is  w/2  and  M  =  yls  wl3. 
Applying  the  loads  in  the  form  of  a  triangle  having  its  apex 


128  REINFORCED  CONCRETE 

at  the  middle  of  the  beam,  the  maximum  moment  will  be,  for 
the  longer  beam 

wl4     bin    _l\ 
~  64  +  I*  4  V2       6/ 

=  At0i&P  (H2) 

and  for  the  shorter  beam     M  =  iV  wzb*l.  (113) 

If  the  slab  be  square,  the  moment  becomes  ^  wl3,  which  is 
slightly  larger  than  just  found  for  the  parabola. 

The  distribution  being  considered  as  uniform,  the  maximum 
moment  for  the  beams  supporting  a  square  slab  is  gV  wZ3,  which 
result  is  in  error  to  an  appreciable  extent.  These  moments  are 
those  due  to  the  loads  on  one  panel  only. 

If  the  beams  supporting  the  slab  be  continuous  the  above 
moments  will  be  somewhat  reduced  by  the  amount  of  negative 
moments  at  each  support.  These  negative  moments  may  be 
computed  by  methods  given  in  the  reference  at  the  foot  of 
page  122,  but  this  refinement  is  seldom  attempted.  The  recom- 
mendations of  the  Joint  Committee  will  be  found  on  page  148, 
and  it  will  be  noted  that  triangular  distribution  of  loads  on 
the  beam  is  the  one  advocated.  This  gives  the  highest  factor 
of  safety  of  the  three  methods  just  discussed. 

For  example,  a  five-inch  slab  is  to  be  10'  X  12 ';  the  uniform 
load,  including  its  own  weight,  is  to  be  250  pounds  per  square 
foot.  It  is  required  to  find  the  proper  amount  of  reinforcement 
in  each  direction  in  the  slab  and  the  bending  moments  in  the 
supporting  beams.  If  the  slab,  loaded  with  wi  per  square  foot, 
be  considered  as  simply  supported  on  the  two  longer  beams, 
the  bending  moment  is  J  X  .67  X  250  X  1200  X  12  =  301,500 
Ib.  in.  By  reference  to  Plate  I,  N  is  found  to  be  0.14,  and  p 
must  be  about  .0048.  The  percentage  in  the  other  rods  need 
be  only  sufficient  to  support  w  (1.00  —  .67)  =  82.5  pounds  per 
square  foot,  then  N  is  0.10  and  p  is  .002.  If  the  distribution  of 
loads  to  the  beams  be  according  to  the  ordinates  of  a  triangle, 
the  bending  moments  at  the  middle  of  the  longer  and  shorter 
beams  will  be,  from  (112)  and  (113),  ^  X  250  X  .67  X  120  X 
144  X  12  =  2,900,000  Ib.  in.,  and  TV  X  250  X  .33  X  144  X  100 
X  12  =  1,190,000  Ib.  in.  In  the  distribution  according  to  ordi- 
nates of  a  parabola,  the  two  moments  are  2,720,000  Ib.  in.,  and 


ANALYSIS  OF  STRESSES 


129 


1,115,000  Ib.  in.  respectively.  The  uniform  distribution  would 
make  the  moments  2,225,000  Ib.  in.  and  893,000  Ib.  in.  respect- 
ively. 

Girderless  Floors.  For  several  years  floors  have  been  built 
after  a  design  that  is  simply  a  thin  slab  resting  upon  columns, 
but  the  beam  and  girder  construction  has  been  much  more 
common.  The  pioneer  in  the  field  of  girderless  design  has 
disclaimed  any  very  theoretical  analysis  of  stresses  *  in  the  slabs, 
but  has  built  many  floors  of  this  kind  that  are  apparently  satis- 


Diana. 
Diam. 
54  Diam. 


FIG.  58. 

factory.  It  is  probable  that  the  lack  of  certainty  as  to  the 
stresses  resulting  from  loads  on  this  type  of  floor  has  made 
other  designs  the  more  common.  Many  tests  of  loaded  floors 
have  been  made  from  time  to  time  to  determine  the  deflections 

1  See  Trans.  Am.  Soc.  C.  E.,  Vol.  LVI,  1906,  pp.  297-309. 


130 


REINFORCED  CONCRETE 


and  whether  or  not  dangerous  cracks  appear,  but  the  test  alluded 
to  below  is  of  special  interest  and  reliability. 

The  test1  was  made  on  eight  adjacent  panels  over  an  area 
three  panels  square,  one  corner  panel  being  omitted.  The 
dimensions  of  the  panels  were  18  feet,  8  inches  by  19  feet,  1  inch, 
and  the  slab  was  9A  inches  thick.  The  concrete  was  good 
quality  1-2-4  mixture,  and  the  floor  was  designed  to  carry 
a  uniform  load  of  225  pounds  per  square  foot.  During  the 
test  the  applied  load  was  350  pounds  per  square  foot.  The 
amount  and  distribution  of  the  metal  reinforcement  are  shown 
in  Fig.  58.  It  is  seen  that  the  rods  are  arranged  in  bands  between 
the  columns  each  way,  and  also  on  both  diagonals  of  the  panels. 
In  this  manner  the  entire  slab  is  covered,  and,  around  the  column 
heads,  the  three  systems  overlap.  In  addition,  eight  radial 
IJ-inch  rods  are  inserted  over  the  columns.  The  summary  of 
observed  stresses  is  given  in  the  following  table: 

SUMMARY  OF  STRESSES 


Load,  225  Ib.  sq.  ft.  . 

Load,  350  Ib.  sq.  ft. 

Live 
load 

Dead 
load 

Total 

Live 
load 

Dead 
load 

Total 

Steel  overhead 

T^.           ,  ,       ,   (  Maximum 
D,agonalband|Average 

13,800 
11,000 

6900 
5500 

20,700 
16,500 

24,200 
18,800 

6900 

5500 

31,100 
24,300 

p        ,       ,  (  Maximum 

10,000 

5000 

15,000 

18,800 

5000 

23,800 

an    (  Average 

9,000 

4500 

13,500 

17,200 

4500 

21,700 

Steel  at  Center 

T^.           1  1      j     (  Maximum 
Diagonal  band    j  Average 

2,400 
2,000 

1200 
1000 

3,600 
3,000 

4,800 
4,800 

1200 
1000 

6,000 
5,800 

Cross  band  {  M;™ 

2,800 
2,500 

1400 
1300 

4,200 
3,800 

8,000 
6,600 

1400 
1300 

9,400 
7,900 

Outer  pane,s{M™ 

4,600 
3,800 

2300 
1900 

6,900 
5,700 

10,400 
8,000 

2300 
1900 

12,700 
9,900 

Concrete  at  Capital 

Diagonal  (  Maximum 

530 

265 

795 

800 

265 

1,065 

direction  (  Average 

500 

250 

750 

750 

250 

1,000 

Cross  direction   {  ^™™m 

500 
468 

250 
234 

750 
700 

800 
750 

250 
234 

1,050 
984 

The  author  of  the  paper  quoted  concludes  that  the  test  indi- 
cates that  the  moments  are  much  greater  at  the  support  than 


By  Arthur  R.  Lord,  described  in  Engineering  Record,  January,  1911. 


ANALYSIS  OF  STRESSES  131 

at  the  center  of  the  span,  that  the  latter  stresses  are  much  smaller 
than  usually  computed,  and  that  the  steel  at  the  center  receives 
its  maximum  stress  when  one  panel  only  is  loaded.  The  points 
of  maximum  negative  moment  were  found  to  be  nearly  over 
the  edges  of  the  capitals  of  the  columns,  or  about  0.2  L  from 
the  middle  of  the  column  head,  L  being  the  length  of  the  side 
of  a  panel.  The  points  of  inflection  are  about  0.55  L  apart, 
or  about  0.225  L  from  the  middle  of  the  projection  of  the  column. 
The  size  of  the  capital  of  the  column  is  of  vital  importance  and 
a  diameter  of  less  than  0.2  L  will  cause  excessive  shearing  and 
flexural  stresses. 

PROBLEMS 

58.  A  4-inch  slab  is  12  feet  by  15  feet,  and  carries  a  uniform  load  of 
200  pounds  per  square  foot.     Find  the  reinforcement  necessary  in  the 
slab  in  both  directions,  and  also  find  the  bending  moment  in  the  sup- 
porting beams. 

59.  Deduce  the  positive  and  negative  bending  moments  in  a  contin- 
uous beam  of  four  equal  spans  if  the  loads  be  distributed  as  the  ordinates 
of  a  triangle. 

60.  Read  the  article  by  Mr.  P.  E.  Stevens,  Trans.  Am.  Soc.  C.  E., 
Vol.  LX,  1908,  on  moments  in  continuous  beams  of  non-uniform  moments 
of  inertia. 

61.  A  square  slab  is  to  support  a  load  of  175  pounds  per  square  foot 
of  area.     If  the  span  in  each  direction  be  18  feet,  and  the  whole  amount 
of  reinforcement  be  one  per  cent.,  what  should  be  the  thickness  of  the 
slab  from  top  to  bottom? 

62.  What  uniform  load  per  square  foot  can  be  safely  carried  by  a 
square  slab  15  feet  on  a  side,  if  the  depth  be  4  inches  and  the  reinforce- 
ment be  0.8  per  cent.? 

63.  What  should  be  the  dimensions  of  the  supporting  beams  in  the 
last  two  problems  if  the  reinforcement  be  1  per  cent,  and  the  depth 
equal  to  two  times  the  breadth? 


CHAPTER  V 
ELEMENTARY  DESIGN 

Economical  Proportions  of  Beams.  As  reinforced  concrete 
beams  are  composed  of  material  of  different  cost,  and  as  the 
breadth  and  depth  are  involved  to  different  extents  in  making 
up  the  strength,  the  cost  of  beams  of  the  same  strength  may  vary 
considerably  with  different  designs. 

value  of  steel  per  unit  of  volume 
value  of  concrete  per  unit  of  volume 

v    =  value  of  steel  per  unit  of  volume 
vf  =  value  of  concrete  per  unit  of  volume 
V  =  value  of  the  beam  per  unit  of  length 
Then  V  =  bdv'  +  Av 

(a)  When  the  breadth  is  constant,  a  certain  breadth  is  required 
to  securely  bond  the  reinforcement.  From  (15),  (7)  and  (8) 


,      _  _n  __    /6  (r  +  n)*b  M 
h  +  n  V  n 


2r  (r  +  n       nC  (3  r  +  2  n) 


+ 


2n       2r\/3r  +  2n 
3r  +  n  9r  +  4n 

17  :~  (3  r  +  2  n)|       n  q  2  r2  (3  r  +  2  n)} 

Putting  this  equal  to  zero  and  simplifying, 


3Q 
40 

Dividing  by  n3 

r3,     r2         3  g 


This  equation  is  platted  in  Plate  XII  with  r/n  as  ordinates,  and 
q/n  as  abscissas.     The  solution  gives  approximately 

q  (116) 


132 


ELEMENTARY  DESIGN  133 

This  is  the  value  of  r  necessary  to  make  V  in  equation  (114)  a 
minimum.  When  r  is  known,  p  and  d  are  readily  found  from 
Plate  I. 

(6)  When  the  Depth  is  Constant.     From  formula  (15)  and  (7) 
a  value  of  V  is  found  which  contains  d  instead  of  6  as  above. 

6  Mv'  (  (r  +  nY  (n  +  r)nq    \ 

Cdn    \3  r  +  2  n  ^  2  r  (3  r  +  2  n)/ 

(r  +  n)  (3  r  +  2  n)  -  3  (r  +  n)2 


dr   '     Cdn   \  (3  r  +  2  n)2 

qnr  (3  r  +  2  n)  -  2  nff  (r  +  n)  (3  r  +  n) 
2r2(3r  +  2  n)2 

Putting  this  equal  to  zero  and  reducing  gives 

r4  +  77  nr*  +  f  ^-  -  ^ )  r2  -  n2  or  -  7:  n3  o  =  0 


Dividing  by  n4  (118) 


_  ..          =  0 

r  3  W        \3       2  ny  n2       n  n        3  n 

In  Plate  XII  the  curve  representing  this  equation  is  marked 
"  depth  constant."  For  values  of  q  and  n  in  common  use,  the 
curve  is  approximately  given  by 

r  =  .8  Vnq  (119) 

From  this  curve,  and  approximately  from  (118),  the  value 
of  r  that  makes  V  in  (117)  a  minimum  is  found.  Then  the 
beam  may  be  designed  from  Plate  I.  Since  the  head  room  is 
often  fixed,  it  is  a  common  problem  to  design  beams  with  a  given 
depth. 

(c)  When  the  Area  of  the  Cross-section  is  Constant.  It  may 
happen  that  the  area,  as  determined  by  (a)  or  (6),  will  not  be 
sufficiently  great  to  insure  the  necessary  shearing  strength  which 
is  the  unit  shearing  strength  multiplied  by  bd.  For  any  given 
problem  bd  is,  then,  a  constant  which  is  known  and  which  may 
be  substituted  in  (16),  giving  an  equation  with  Ni  and  d  as  vari- 
ables. Plate  I  shows  that  the  percentage  of  steel  decreases 
with  Nij  so  d  should  be  made  large  for  minimum  cost.  If  either 
the  breadth  or  the  depth  be  otherwise  limited,  the  necessary 


134  REINFORCED  CONCRETE 

area  of  cross-section  must  be  provided  for  by  a  variation  in  the 
other  dimension. 

The  cross-section  is,  however,  seldom  fixed  by  the  shearing 
stresses  except  near  the  end  of  the  beam. 

(d)  When  the  Ratio  of  Breadth  to  Depth  is  Fixed.  This  condi- 
tion is  sometimes  imposed  to  preserve  symmetry  and  for  other 
reasons.  From  (15)  there  may  be  deduced 


yd.  =  &/ 
d  \ 


,Cn  (3  r  -f  2  n), 
and 

-636M2Y    /    (r  +  n)*  <?n  (r  +  n)*  \ 

2n2  y      V(3  r  +  2  n)I  "*"  2  r  (r  +  n)  (3  r  +  2  n)  V 

By  putting  the  first  derivative  equal  to  zero  there  results 

^+3r?  +  3n2~nn2~l2nn~2n  = 

This  curve  is  platted  on  Plate  XII,  and  from  it  the  value  of  r 
is  found  that  will  make  the  cost  a  minimum.  The  value  of  r  is 
also  given  approximately  by  the  equation 

r  =  l.OSVng  (122) 

Of  the  three  curves  thus  far  platted  on  Plate  XII  that  repre- 
senting a  constant  breadth  is  seen  to  show  the  most  rapid  change 
in  r  for  a  given  change  in  q. 

(e)  When  the  Breadth  is  Constant  and  the  Stress  is  Ultimate. 
Using  the  parabolic  relation,  formula  (38)  may  be  changed  to 

/3  Mb\*      r  +  2  n 
hfj  —  f . 

~\Cn)    (4  r  +  5  n)* 
and 


,(4r  +  5n)*    '   3  r  (4  r  +  5  n 

Then  the  first  derivitive  of  V  with  respect  to  r  equated  to  zero 
gives 


n3       2  n2         nn       3  n 

A  part  of  this  curve  is  shown  in  Plate  XI  and  may  be  extended 
by  the  approximate  value  of 

r  =  2.07V™?  (125) 

but  r  is  seldom  more  than  4  n. 


ELEMENTARY  DESIGN 


135 


Values  of  3  -r-  n 


\ 


PLATE  XII. 


136  REINFORCED  CONCRETE 

The  above  discussion  involves  but  few  phases  of  the  subject 
of  economic  design  and  is  given  more  to  indicate  methods  than 
to  designate  results.  The  analysis  may  be  extended  to  include 
the  weight  of  the  beam  itself  but  it  is  usual  to  allow  for  a  dead 
load  of  about  one-half  the  line  load. 

In  a  T-beam  there  are  four  dimensions  and  the  amount  of 
steel  to  be  determined  and,  as  a  result,  formulas  become  un wieldly 
unless  some  assumptions  are  made.  If  the  flange  be  designed 
separately  to  carry  the  load  between  beams,  the  design  of  the 
T-beam  will  be  that  of  the  web  and  the  flange  remains  constant. 

For  example,  let  it  be  required  to  design  a  rectangular  beam 
15  feet  long  to  have  a  safe  bending  moment  of  500,000  Ib.  in. 
Let  the  price  of  the  steel  be  3J  cents  per  pound  and  that  of  the 
concrete  $8  per  cubic  yard,  and  let  the  breadth  be  fixed  at 
15  inches.  Plate  XIII  gives,  for  these  prices,  58  for  the  value 
of  q.  Or,  if  n  be  15,  q/n  =  3.87.  With  this  value  of  q/n,  Plate 
XII  gives  2.44  for  r/n  and  r  =  36.6,  and  Plate  I,  in  turn,  gives 
NI  =  .132  and  p  =  A  per  cent,  of  reinforcement.  Then,  132  X 
600  X  15  X  d?  =  500,000  and  d  =  20.5  inches.  The  cost  of  the 
beams  will  be  15  X  20.5  X  15  -r-  3888  X  8  =  $9.49  for  the  con- 
crete and  9.47  X  .004  X  58  =  $2.20  for  the  steel,  or  $11.69 
total.  This  does  not  include  the  inch  and  a  half  of  depth  below 
the  reinforcement  as  this  is  constant  whatever  the  design  may  be. 

If  the  depth  be  taken  as  19  inches,  NI  =  500,000  -r-  (15  X  192 
X  600)  =  .154,  r  =  28  and  p  =  .0062.  Then  the  price  will  be, 
for  the  beam,  15  X  19  X  15  -5-  3888  X  8  (1  +  .0062  X  58)  = 
$11.95.  If  the  depth  be  assumed  as  21  inches  Ni  =  500,000  -f- 
(15  X  212  X  600)  =  .126  and  p  =  .00355.  Then  the  cost  of  the 
beam  is  15  X  21  X  15  4-  3888  X  8(1  +  .00355  X  58)  =  $11.76. 
Both  these  prices  are  seen  to  be  but  slightly  greater  than  that 
deduced  by  (a). 

As  an  example  under  (6)  let  it  be  necessary  to  fix  the  depth  in 
the  above  beam  at  18  inches.  It  is  required  to  find  the  most 
economical  breadth.  From  Plate  XII  r  is  found  to  be  27.5  and 
Plate  I  gives  .156  X  600  X  324  6  =  500,000,  from  which  b  = 
16.5  inches  and  p  is  .0065.  The  beam  will  cost  8  X  16.5  X  18  X 
15  -r-  3888  (1  +  .0065  X  58)  =  $12.61.  For  b  =  15  inches  the 
cost  is  $12.67  and  for  b  =  18  this  is  increased  to  $12.95. 

The  shear  in  the  above  examples  is  J  W=  11,111  pounds  and 
for  b  =  15  and  d  =  20,  v  is  37  pounds  per  square  inch.  This  is 


ELEMENTARY  DESIGN 


137 


PLATE  XIII. 


1:1  INKOIU  I  I)  ('<>N<  1:1  II 

nut    e\ce     i\  e  even   1  1    i  HUH-  of  I  In-  rods  l>r  hen  I    up  as  they   ne;irl\ 
al\\a\       .iir         Mrrr    (lie    shear    i       nut     ;i    controlling    factor    in    I  he 


As  :in  example  of  design  when  <l  l>  is  fixed  let  (his  r;itiu  he  1?.."), 
\vitli  Hie  length  o!'  heam,  the  hendini;  moment,  and  the  prices  :is 
helore.  Krom  IMale  Ml  /•  is  found  to  he  .'JO.S.  Plate  I  j;i\es  A, 

M  .146  and  p  *s  .00532.    The.,    I  ir.   -  r.oo   -   r,  2561      ..oo.ooo, 

an.  I  /i         '.».'/   incher,  and  «/         'Jl.:i  Indies.      'I'ln-  cost   of  this  Ix-ain 

in  9.7  X  2-U  X   l-r>  X  8  +  3888  (1  +  .00,r>;i2  X  58)  -  $«,»..  vj. 

\\  h.-n  /'  i     '»  inclu'H   Ar,  is     IS!?  and  f>  is  .0107  and  Ihr  ro-t   l.rroiurs 

0  X  22.f>  X  15  X  8  H-  3888    (1  -|-  .0107  X  58)  •    111.30,      li   h 

\\.n    as.sunu-d  to  IM-   II   inches    A,  \\ould  I  M-  .O'U   and  />  would  he 
.0010.      Then    the    value    umild    he    II    \   'J7..'i   X   15  X  8  -r   liSSS 

(1  -f  58  X  .0016)  -  $10.'" 

II    </    h    he    Jissillne.i  !()     .     I.)    Ihe    deduced    \;ihle    of    Ilie    \\ldlh 

\\ill   he    IT)  iiu-lifs  as  in   Ihe  example  under   (11). 

PROBLEMS 

HI  If  (he  hiv.Millli  of  :.  heam  he  I  L>  inches,  ('  he  .MM)  Ih.  M,  in, 
slcrl  he  ;i  cents  per  pound,  ami  r.Micrele  Sli  |HT  cuhic  \:ir«l,  \\liat  \\ill 
he  (he  least  CIK-I  ••!  .1  I  ifel\  \\illi  Lind  a  l-.endinu,  nioincnt  of 

HXI.OOO  II.  m  " 

!'».")  Deduce  formula-,  foi  economic  dinu-nMons  »>f  hcains,  (akin.i;  inlo 
account  the  weight  o!"  the  hcani  it  rll 

C.(i  A  hcain  \\as  dcsii;nt>d  to  cans  a  load  caiiMicj,  a  hcndmu  moment 
of  ,'i()0,(MX)  pounds  inches,  ('  was  .M)0  Ih  s(j.  in  .,  /»  \vas  10  inches,  and 

\->  is  'J'J  indues.  If  \\\v  cost,  of  sttM'l  Were  '.'  ,'  ccnl>  pel  pound,  \\hat  \\as 
lli.  co  I  of  the  concrete,  if  the  design  \\crc  economical  and  the  depth 
\\erc  lixcd"  \\hat  u  hen  the  htcadth  and  \\hcn  l>  ./  \\ci'c  lived'.' 

(>7.  A  heam  is  I'.'  inches  h\  '.'  I  inches  and  hi  feel  span  With  /•  «/ 
li\.>l,  ('  -  ,r)(K)  Ih.  s(j.  in.,  steel  al  a  ccnls  per  pound,  and  concrete 
lit  $7  prr  cuhic  \ard,  \\hat  pet  cent  a^c  of  Meel  can  he  ust>d  \\ith  economy, 
Htul  what  will  he  the  hendin^  moment'.' 


UKPOHT   OK   Till     JOINT    roMMIT'lTl 

Ar  Ihr  annual  convention  of  the  American  S.>det\  of  Civil 
Kni;inrrrs  lirld  al  Asheville,  N.  ('.,  .luno  11,  ItHKl,  the  folloxx  inj; 
resolution  was  adopted 

It  is  the  sense  of  this  meetinu;  that  a  special  committee  he  appointed 
to  lake  up  the  question  of  concrete  and  steel  concrete,  and  that  such 
committee  cooperate  \\ith  the  American  Socid  \  lor  TeMmii  Materials 


i  i  i  MI  \  r\i{\    OKSKJN 

ami  tin-  American  Railua)    Lngmecring  ami  Maintenance  of  Way  Asso- 
ciation. 

Following  the  adoption  of  this  resolution,  a  special  committee 
on  concrete  and  steel  concrete  \\as  appointed  by  the  Board  of 
Direction  on  Max  ;U ,  191)4.  At  the  annual  nu>rt  ing,  held  Jan- 
u.i!  \  IS,  100,~>,  the  title  of  this  special  committee  was,  at  the  re- 
quest of  the  committee,  changed  to  "Special  Committee  on 
Concrete  and  Reinforced  Concrete." 

At  the  annual  meeting  of  the  American  Society  for  Testing 
Material  held  ,luly  1,  lOltt,  at  the  Delaware  Water  Gap,  the 
following  resolution  A\as  unanimously  adopted: 

That   the  Executive  Committee  be  requested  to  consider  the  desira- 

hility  of  appointing  a  committee  on  "  Reinforced  Concrete,"  with  a 
\ie\\  of  cooperating  with  the  committees  of  other  societies  in  the  study 

of  the   Mihject. 

At  the  meeting  of  the  Executive  Committee  of  this  Society, 
held  December  5,  I'.Htt,  a  .special  committee  on  "Reinforced 
( 'oncrete"  was  a]>i>ointed. 

The  American  Railway  Engineering  and  Maintenance  of  Way 
As>ociati»»n  appointed  a  Committee  on  Masonry  on  July  20,  1809, 
with  instruction  as  a  part  of  its  duties  to  ]>repare  specifications 
for  concrete  masonry.  A  preliminary  set  of  specifications  for 
Portland  cement  concrete  was  reported  to  and  adopted  by  the 
Association  on  March  10,  1003.  At  the  meeting  held  in  Chicago 
on  March  17,  1004,  the  Committee  on  Masonry  was  authorized 
to  cooperate  \\ith  the  special  committee  on  Concrete  and  Re- 
inforced Corn-ret e  of  the  American  Society  of  Civil  Engineers, 
and  following  this  action  a  special  sub-committee  was  appointed. 

At  a  meeting  of  the  several  special  committees  representing 
the  :d»o\e  mentioned  Societies,  and  also  a  committee  appointed 

by  the  Association  of  American  Portland  Cement  Manufacturers, 
held  at  Atlantic  City,  N.  J.,  June  17,  1004,  arrangements  were 
completed  for  collaborating  the  \\ork  of  these  several  committees 
through  the  formation  of  the  joint  committee  on  Concrete  and 
Reinforced  ( 'oncrete. 

Some  of  the  specifications  recommended  by  this  joint  com- 
mittee have  been  quoted  in  Chapter  II.  The  following  quota- 
tions are  inserted  here  to  furnish  a  guide  for  class-room  design 


140  REINFORCED  CONCRETE 

and  also  because  they  represent  standard  practice  among  Ameri- 
can engineers. 

PREPARATION  AND  PLACING  OF  MORTAR  AND  CONCRETE 

1.  Proportions.  The  materials  to  be  used  in  concrete  should 
be  carefully  selected,  of  uniform  quality,  and  proportioned  with 
a  view  to  securing  as  nearly  as  possible  a  maximum  density. 

(a)  Unit  of  Measure.  —  The  unit  of  measure  should  be  the 
barrel,  which  should  be  taken  as  containing  3.8  cubic  feet.  Four 
bags  containing  94  pounds  of  cement  each  should  be  considered 
the  equivalent  of  one  barrel.  Fine  and  coarse  aggregate  should 
be  measured  separately  as  loosely  thrown  into  the  measuring 
receptacle. 

(6)  Relation  of  Fine  and  Coarse  Aggregate.  —  The  fine  and 
coarse  aggregate  should  be  used  in  such  relative  proportions  as 
will  insure  maximum  density.  In  unimportant  work  it  is  suffi- 
cient to  do  this  by  individual  judgment,  using  correspondingly 
higher  proportions  of  cement;  for  important  work  these  propor- 
tions should  be  carefully  determined  by  density  experiments 
and  the  sizing  of  the  fine  and  coarse  aggregates  should  be  uni- 
formly maintained  or  the  proportions  changed  to  meet  the  vary- 
ing sizes. 

(c)  Relation  of  Cement  and  Aggregates.  —  For  reinforced  con- 
crete construction  a  density  proportion  based  on  1 : 6  should 
generally  be  used,  i.e.,  one  part  of  cement  to  a  total  of  six  parts 
of  fine  and  coarse  aggregates  measured  separately. 

In  columns  richer  mixtures  are  often  required,  while  for  mas- 
sive masonry  or  rubble  concrete  a  leaner  mixture  of  1 :  9  or  even 
1 :  12  may  be  used.  These  proportions  should  be  determined 
by  the  strength  or  wearing  qualities  required  in  the  construction 
at  the  critical  period  of  its  use.  Experienced  judgment  based  on 
individual  observation  and  tests  of  similar  conditions  in  similiar 
localities  is  the  best  guide  as  to  the  proper  proportions  for  any 
particular  case. 

2.  Mixing.  The  ingredients  of  concrete  should  be  thoroughly 
mixed  to  the  desired  consistency,  and  the  mixing  should  continue 
until  the  cement  is  uniformly  distributed  and  the  mass  is  uniform 
in  color  and  homogeneous,  since  maximum  density  and  therefore 
greatest  strength  of  a  given  mixture  depend  largely  on  thorough 
and  complete  mixing. 


ELEMENTARY  DESIGN  141 

(a)  Measuring  Ingredients.  —  Methods  of  measurements  of 
the  proportions  of  the  various  ingredients,  including  the  water, 
should  be  used,  which  will  secure  separate  uniform  measurements 
at  all  times. 

(6)  Machine  Mixing.  —  When  the  conditions  will  permit,  a 
machine  mixer  of  a  type  which  insures  the  uniform  proportion- 
ing of  the  materials  throughout  the  mass  should  be  used,  since  a 
more  thorough  and  uniform  consistency  can  be  thus  obtained. 

(c)  Hand  Mixing.  —  When  it  is  necessary  to  mix  by  hand, 
the  mixing  should  be  on  a  water-tight  platform  and  especial 
precautions  should  be  taken  to  turn  the  materials  until  they  are 
homogeneous  in  appearance  and  color. 

(d)  Consistency.  —  The  materials  should  be  mixed  wet  enough 
to  produce  a  concrete  of  such  a  consistency  as  will  flow  into  the 
forms  and  about  the  metal  reinforcement,  and  which,  at  the  same 
time,  can  be  conveyed  from  the  mixer  to  the  forms  without 
separation  of  the  coarse  aggregate  from  the  mortar. 

(e)  Retempering.  —  Retempering    mortar    or    concrete,    i.e., 
remixing  with  water  after  it  has  partially  set,  should  not  be  per- 
mitted. 

3.  Placing  of  Concrete,  (a)  Methods.  —  Concrete  after  the 
addition  of  water  to  the  mix  should  be  handled  rapidly,  and  in 
as  small  masses  as  is  practicable,  from  the  place  of  mixing  to  the 
place  of  final  deposit,  and  under  no  circumstances  should  con- 
crete be  used  that  has  partially  set  before  final  placing.  A  slow- 
setting  cement  should  be  used  when  a  long  time  is  likely  to  occur 
between  mixing  and  final  placing. 

The  concrete  should  be  deposited  in  such  a  manner  as  will 
permit  the  most  thorough  compacting,  such  as  can  be  obtained 
by  working  with  a  straight  shovel  or  slicing  tool  kept  moving  up 
and  down  until  all  the  ingredients  have  settled  in  their  proper 
place  by  gravity  and  the  surplus  water  has  been  forced  to  the 
surface. 

In  depositing  the  concrete  under  water,  special  care  should 
be  exercised  to  prevent  the  cement  from  being  floated  away, 
and  to  prevent  the  formation  of  laitance,  which  hardens  very 
slowly  and  forms  a  poor  surface  on  which  to  deposit  fresh  con- 
crete. Laitance  is  formed  in  both  still  and  running  water,  and 
should  be  removed  before  placing  fresh  concrete. 

Before  placing  the  concrete,  caje  should  be  taken  to  see  that 


142  REINFORCED  CONCRETE 

the  forms  are  substantial  and  thoroughly  wetted  and  the  space 
to  be  occupied  by  the  concrete  is  free  from  debris.  When  the 
placing  of  the  concrete  is  suspended,  all  necessary  grooves  for 
joining  future  work  should  be  made  before  the  concrete  has  had 
time  to  set. 

When  work  is  resumed,  concrete  previously  placed  should  be 
roughened,  thoroughly  cleansed  of  foreign  material  and  laitance, 
drenched  and  slushed  with  a  mortar  consisting  of  one  part 
Portland  cement  and  not  more  than  two  parts  fine  aggregate. 

The  faces  of  concrete  exposed  to  premature  drying  should  be 
kept  wet  for  a  period  of  at  least  seven  days. 

(b)  Freezing    Weather.  —  Concrete    for    reinforced    structures 
should  not  be  mixed  or  deposited  at  a  freezing  temperature, 
unless  special  precautions  are  taken  to  avoid  the  use  of  materials 
containing  frost  or  covered  with  ice  crystals,  and  to  provide 
means  to  prevent  the  concrete  from  freezing  after  being  placed 
in  position  and  until  it  has  thoroughly  hardened. 

(c)  Rubble  Concrete.  —  Where  the  concrete  is  to  be  deposited 
in  massive  work,  its  value  may  be  improved  and  its  cost  materi- 
ally reduced  through  the  use  of  clean  stones  thoroughly  embedded 
in  the  concrete  as  near  together  as  is  possible  and  still  entirely 
surrounded  by  concrete. 

4.  Forms.  Forms  should  be  substantial  and  unyielding,  so 
that  the  concrete  shall  conform  to  the  designed  dimensions  and 
contours,  and  should  be  tight  to  prevent  the  leakage  of  mortar. 

The  time  for  removal  of  forms  is  one  of  the  most  important 
steps  in  the  erection  of  a  structure  of  concrete  or  reinforced  con- 
crete. Care  should  be  taken  to  inspect  the  concrete  and  ascer- 
tain its  hardness  before  removing  the  forms. 

So  many  conditions  affect  the  hardening  of  concrete  that  the 
proper  time  for  the  removal  of  the  forms  should  be  decided  by 
some  competent  and  responsible  person,  especially  where  the 
atmospheric  conditions  are  unfavorable. 

DETAILS  OF  CONSTRUCTION 

1.  Joints,  (a)  Reinforcement.  —  Wherever  in  tension  rein- 
forcement it  is  necessary  to  splice  the  reinforcing  bars,  the 
length  of  lap  shall  be  determined  on  the  basis  of  the  safe  bond 
stress  and  the  stress  in  the  bar  at  the  point  of  splice;  or  a  con- 
nection shall  be  made  between  ,the  bars  of  sufficient  strength  to 


ELEMENTARY  DESIGN  143 

carry  the  stress.  Splices  at  points  of  maximum  stress  should 
be  avoided.  In  columns  large  bars  should  be  properly  butted 
and  spliced;  small  bars  may  be  treated  as  indicated  for  tension 
reinforcement  or  their  stress  may  be  taken  off  by  being  imbedded 
in  large  masses  of  concrete.  At  foundations,  bearing  plates 
should  be  provided  for  large  bars  or  structural  forms. 

(b)  Concrete.  —  For  concrete  construction  it  is  desirable  to 
cast  the  entire  structure  at  one  operation,  but  as  this  is  not 
always  possible,  especially  in  large  structures,  it  is  necessary  to 
stop  the  work  at  some  convenient  point.  This  point  should  be 
selected  so  that  the  resulting  joint  may  have  the  least  possible 
effect  on  the  strength  of  the  structure.  It  is  therefore  recom- 
mended that  the  joint  in  columns  be  made  flush  with  the  lower 
side  of  the  girders;  that  the  joints  in  girders  be  at  a  point  mid- 
way between  supports,  but  should  a  beam  intersect  a  girder  at 
this  point,  the  joint  should  be  offset  a  distance  equal  to  twice 
the  width  of  the  beam;  that  the  joints  in  the  members  of  a  floor 
system  should  in  general  be  made  at  or  near  the  center  of  the 
span. 

Joints  in  columns  should  be  perpendicular  to  the  axis  of  the 
column,  and  in  girders,  beams,  and  floor  slabs  perpendicular  to 
the  plane  of  their  surfaces. 

2.  Shrinkage.     Girders    should   never   be    constructed   over 
freshly  formed  columns  without  permitting  a  period  of  at  least 
two  hours  to  elapse,  thus  providing  for  settlement  or  shrinkage 
in  the  columns.     Before  resuming  work,  the  top  of  the  column 
should  be  thoroughly  cleansed  of  foreign  matter  and  laitance. 
If  the  concrete  in  the  column  has  become  hard,  the  top  should 
also  be  drenched  and  slushed  with  a  mortar  consisting  of  one 
part  Portland  cement  and  not  more  than  two  parts  fine  aggre- 
gate before  placing  additional  concrete. 

3.  Temperature  Changes.     Concrete  is  sensitive  to  tempera- 
ture changes,  and  it  is  necessary  to  take  this  fact  into  account 
in  designing  and  erecting  concrete  structures.     In  some  posi- 
tions the  concrete  is  subjected  to  a  much  greater  fluctuation  in 
temperature  than  in  others,  and  in  such  cases  joints  are  neces- 
sary.    The  frequency  of  these  joints  will  depend,  first  upon  the 
range    of   temperature   to   which    concrete   will   be    subjected; 
second,  upon  the  quantity  and  position  of  the  reinforcement. 
These  points  should  be  determined  and  provided  for  in  the 


144  REINFORCED  CONCRETE 

design.  In  massive  work,  such  as  retaining  walls,  abutments, 
etc.,  built  without  reinforcement,  joints  should  be  provided, 
approximately,  every  50  ft.  throughout  the  length  of  the  struc- 
ture. To  provide  against  the  structures  being  thrown  out  of 
line  by  unequal  settlement,  each  section  of  the  wall  may  be 
tongued  and  grooved  into  the  adjoining  section.  To  provide 
against  unsightly  cracks,  due  to  unequal  settlement,  a  joint 
should  be  made  at  all  sharp  angles. 

4.  Fire-proofing.  The  actual  fire  tests  of  concrete  and  re- 
inforced concrete  have  been  limited,  but  experience,  together 
with  the  results  of  tests  so  far  made,  indicate  that  concrete  may 
be  safely  used  for  fire-proofing  purposes.  Concrete  itself  is 
incombustible  and  reasonably  proof  against  fire  when  composed 
of  a  siliceous  sand  and  a  hard  coarse  aggregate  such  as  igneous 
rock. 

For  a  fire-proof  covering  these  same  materials  may  be  used 
or  clean  hard  burned  cinders  may  be  substituted  for  the  coarse 
aggregate. 

The  low  rate  of  heat  conductivity  of  concrete  is  one  reason 
for  its  value  for  fire-proofing.  The  dehydration  of  the  water  of 
crystallization  of  concrete  probably  begins  at  about  500°  F.  and 
is  completed  at  about  900°  F.,  but  experience  indicates  that  the 
volatilization  of  the  water  absorbs  heat  from  the  surrounding 
mass,  which,  together  with  the  resistance  of  the  air  cells,  tends 
to  increase  the  heat  resistance  of  the  concrete,  so  that  the  process 
of  dehydration  is  very  much  retarded.  The  concrete  that  is 
actually  affected  by  fire  remains  in  position  and  affords  protec- 
tion to  the  concrete  beneath  it. 

It  is  recommended  that  in  monolithic  concrete  columns,  the 
concrete  to  a  depth  of  one  and  one-half  inches  be  considered  as 
protective  covering  and  not  included  in  the  effective  section. 

The  thickness  of  the  protective  coating  required  depends 
upon  the  probable  duration  of  a  fire  which  is  likely  to  occur  in 
the  structure  and  should  be  based  on  the  rate  of  heat  conduc- 
tivity. The  question  of  the  conductivity  of  concrete  is  one  which 
requires  further  study  and  investigation  before  a  definite  rate 
for  different  classes  of  concrete  can  be  fully  established.  How- 
ever, for  ordinary  conditions  it  is  recommended  that  the  metal 
in  girders  and  columns  be  protected  by  a  minimum  of  two  inches 
of  concrete;  that  the  metal  in  beams  be  protected  by  a  minimum 


ELEMENTARY  DESIGN  145 

of  one  and  one-half  inches  of  concrete,  and  that  the  metal  in 
floor  slabs  be  protected  by  a  minimum  of  one  inch  of  concrete. 

It  is  recommended  that  the  corners  of  columns,  girders,  and 
beams  be  beveled  or  rounded,  as  a  sharp  corner  is  more  seriously 
affected  by  fire  than  a  round  one. 

5.  Waterproofing.     Many    expedients    have    been    used    to 
render  concrete  impervious  to  water  under  normal  conditions, 
and  also  under  pressure  conditions  that  exist  in  reservoirs,  dams, 
and   conduits   of   various   kinds.     Experience  shows,   however, 
that  where  mortar  or  concrete  is  proportioned  to  obtain  the 
greatest  practicable  density  and  is  mixed  to  a  rather  wet  con- 
sistency, the  resulting  mortar  or  concrete  is  impervious  under 
ordinary  conditions.     A  concrete  of  dry  consistency  is  more  or 
less  pervious  to  water,  and  compounds  of  various  kinds  have 
been  mixed  with  the  concrete  or  applied  as  a  wash  to  the  surface 
for  the  purpose  of  making  it  water-tight.     Many  of  these  com- 
pounds are  of  but  temporary  value  and  in  time  lose  their  power 
of  imparting  impermeability  to  concrete. 

In  the  case  of  subways,  long  retaining  walls,  and  reservoirs, 
leakage  cracks  may  be  prevented  by  horizontal  and  vertical 
reinforcement  properly  proportioned  and  located,  provided  the 
concrete  itself  is  impervious. 

Such  reinforcement  distributes  the  stretch  due  to  contraction 
or  settlement  so  that  the  cracks  are  too  minute  to  permit  leakage, 
or  are  soon  closed  by  infiltration  of  silt. 

Asphaltic  or  coal-tar  preparations,  applied  either  as  a  mastic 
or  as  a  coating  on  felt  or  cloth  fabric,  are  used  for  waterproofing, 
and  should  be  proof  against  injury  by  liquids  or  gases. 

6.  Surface  Finish.     Concrete  is  a  material  of  an  individual 
type,  and  should  not  be  used  in  imitation  of  other  structural 
materials.     One  of  the  important  problems  connected  with  the 
use  of  concrete  is  the  character  of  the  finish  of  exposed  surfaces. 
The  finish  of  the  surface  should  be  determined  before  the  con- 
crete is  placed,  and  the  work  conducted  so  as  to  make  possible 
the  finish  desired.     For  many  forms  of  construction  the  natural 
surface  of  the  concrete  is  unobjectionable,  but  frequently  the 
marks  of  the  boards  and  the  flat  dead  surface  are  displeasing, 
and  make  some  special  treatment  desirable.     A  treatment  of 
the  surface  which  removes  the  film  of  mortar  and  brings  the 
coarser  particles  of  the  concrete  into  relief  is  frequently  used  to 


146  REINFORCED  CONCRETE 

remove  the  form  markings,  break  the  monotonous  appearance 
of  the  surface,  and  make  it  more  pleasing.  Plastering  of  surfaces 
should  be  avoided,  for  the  other  methods  of  treatment  are  more 
reliable  and  usually  much  more  satisfactory.  Plastering,  even 
if  carefully  applied,  is  likely  to  peel  off  under  the  action  of  frost 
or  temperature  changes. 

DESIGN 

1.  Massive  Concrete.     In  the  design  of  massive  concrete  or 
plain  concrete,  no  account  should  be  taken  of  the  tensile  strength 
of  the  material,  and  sections  should  usually  be  so  proportioned 
as  to  avoid  tensile  stresses.     This  will  generally  be  accomplished, 
in  the  case  of  rectangular  shapes,  if  the  line  of  pressure  is  kept 
within  the  middle  third  of  the  section,  but  in  very  large  struc- 
tures, such  as  high  masonry  dams,  a  more  exact  analysis  may  be 
required.     Structures   of   massive   concrete   are   able   to   resist 
unbalanced  lateral  forces  by  reason  of  their  weight,  hence  the 
element  of  weight  rather  than  strength  often  determines  the 
design.     A  relatively  cheap  and  weak  concrete  will  therefore 
often  be  suitable  for  massive  concrete  structures.     Owing  to  its 
low  extensibility,  the  contraction  due  to  hardening  and  to  tem- 
perature changes  requires  special  consideration,  and,  except  in 
the  case  of  very  massive  walls,  such  as  dams,  it  is  desirable  to 
provide  joints  at  intervals  to  localize  the  effect  of  such  contrac- 
tion.    The  spacing  of  such  joints  will  depend  upon  the  form  and 
dimensions  of  the  structure  and  its  degree  of  exposure. 

Massive  concrete  is  also  a  suitable  material  for  arches  of 
moderate  span  where  the  conditions  as  to  foundations  are 
favorable. 

2.  Reinforced  Concrete.     By  the  use  of  metal  reinforcement 
to  resist  the  principal,  tensile  stresses,   the  concrete  becomes 
available  for  general  use  in  a  great  variety  of  structures  and 
structural  forms.     This  combination  of  concrete  and  steel  is 
particularly   advantageous  in  beams  where  both   compression 
and  tension  exist;   it  is  also  advantageous  in  the  column,  where 
the  main  stresses  are  compressive,  but  where  cross-bending  may 
exist.     The  theory  of  design  will  therefore  relate  mainly  to  the 
analysis  of  beams  and  columns. 

3.  General  Assumptions,     (a)  Loads.  —  The  loads  or  forces 
to  be  resisted  consist  of : 


ELEMENTARY  DESIGN  147 

1.  The  dead  load,  which  includes  the  weight  of  the  structure 
and  fixed  loads  and  forces. 

2.  The  live  load  or  the  loads  and  forces  which  are  variable. 
The  dynamic  effect  of  the  live  load  will  often  require  considera- 
tion.    Any  allowance  for  the  dynamic  effect  is  preferably  taken 
into  account  by  adding  the  desired  amount  to  the  live  load  or  to 
the  live  load  stresses.     The  working  stresses  hereinafter  recom- 
mended are  intended  to  apply  to  the  equivalent  static  stresses 
so  determined. 

In  the  case  of  high  buildings  the  live  load  on  columns  may  be 
reduced  in  accordance  with  the  usual  practice. 

(6)  Lengths  of  Beams  and  Columns.  —  The  span  length  for 
beams  and  slabs  shall  be  taken  as  the  distance  from  center  to 
center  of  supports,  but  shall  not  be  taken  to  exceed  the  clear 
span  plus  the  depth  of  beam  or  slab.  Brackets  shall  not  be 
considered  as  reducing  the  clear  span  in  the  sense  here  intended. 

The  length  of  columns  shall  be  taken  as  the  maximum  unsup- 
ported length. 

(c)  Internal  Stresses.  —  As  a  basis  for  calculations  relating  to 
the  strength  of  structures,  the  following  assumptions  are  recom- 
mended: 

1.  Calculations  should  be  made  with  reference  to  working 
stresses  and  safe  loads  rather  than  with  reference  to  ultimate 
strength  and  ultimate  loads. 

2.  A    plane    section    before    bending    remains    plane    after 
bending. 

3.  The  modulus  of  elasticity  of  concrete  in  compression  within 
the  usual  limits  of  working  stresses  is  constant.     The  distribu- 
tion of  compressive  stresses  in  beams  is  therefore  rectilinear. 

4.  In   calculating  the   moment   of   resistance   of  beams  the 
tensile  stresses  in  the  concrete  shall  be  neglected. 

5.  Perfect   adhesion  is   assumed   between   concrete   and   re- 
inforcement.    Under  compressive  stresses  the  two  materials  are 
therefore  stressed  in  proportion  to  their  moduli  of  elasticity. 

6.  The  ratio  of  the  modulus  of  elasticity  of  steel  to  the  modu- 
lus of  elasticity  of  concrete  may  be  taken  at  15. 

7.  Initial  stress  in  the  reinforcement  due  to  contraction  or 
expansion  in  the  concrete  may  be  neglected. 

It  is  appreciated  that  the  assumptions  herein  given  are  not 
entirely  borne  out  by  experimental  data.     They  are  given  in  the 


148  REINFORCED  CONCRETE 

interest  of  simplicity  and  uniformity,  and  variations  from  exact 
conditions  are  taken  into  account  in  the  selection  of  formulas 
and  working  stresses. 

For  calculations  relative  to  deflections  the  tensile  strength  of 
the  concrete  should  be  taken  into  account.  For  such  calcula- 
tions, also,  a  value  of  8  to  12  for  the  ratio  of  the  moduli  corre- 
sponds more  nearly  to  the  actual  conditions  and  may  well  be 
used. 

4.  Tee  Beams.     In  beam  and  slab  construction,  an  effective 
bond  should  be  provided  at  the  junction  of  the  beam  and  slab. 
When  the  principal,  slab  reinforcement  is  parallel  to  the  beam, 
transverse   reinforcement   should  be  used,   extending   over   the 
beam  and  well  into  the  slab. 

Where  adequate  bond  between  slab  and  web  of  beam  is  pro- 
vided, the  slab  may  be  considered  as  an  integral  part  of  the 
beam,  but  its  effective  width  shall  be  determined  by  the  following 
rules : 

(a)  It  shall  not  exceed  one-fourth  of  the  span  length  of  the 
beam; 

(b)  Its  overhanging  width  on  either  side  of  the  web  shall  not 
exceed  4  times  the  thickness  of  the  slab. 

In  the  design  of  tee  beams  acting  as  continuous  beams,  due 
consideration  should  be  given  to  the  compressive  stresses  at  the 
support. 

5.  Floor  Slabs.     Floor  slabs  should  be  designed  and  reinforced 
as   continuous  over  the   supports.     If   the   length   of  the   slab 
exceeds  1.5  times  its  width  the  entire  load  should  be  carried  by 
transverse  reinforcement.     Square  slabs  may  well  be  reinforced 
in  both  directions.1 

The  loads  carried  to  beams  by  slabs  which  are  reinforced  in 
two  directions  will  not  be  uniformly  distributed  to  the  support- 
ing beam,  and  may  be  assumed  to  vary  in  accordance  with  the 

1  The  exact  distribution  of  load  on  square  and  rectangular  slabs,  supported 
on  four  sides  and  reinforced  in  both  directions,  cannot  readily  be  determined. 
The  following  method  of  calculation  is  recognized  to  be  faulty,  but  it  is  offered 
as  a  tentative  method  which  will  give  results  on  the  safe  side.  The  distribution 
of  load  is  first  to  be  determined  by  the  formula 

ft 

~  ft  4-  64 
in  which  r  =  proportion  of  load  carried  by  the  transverse  reinforcement, 


ELEMENTARY  DESIGN  149 

ordinates  of  a  triangle.  The  moments  in  the  beams  should  be 
calculated  accordingly. 

6.  Continuous  Beams  and  Slabs.     When  the  beam  or  slab 
is  continuous  over  its  supports,  reinforcement  should  be  fully 
provided  at  points  of  negative  moment.     In  computing  the  posi- 
tive and  negative  moments  in  beams  and  slabs  continuous  over 
several  supports,  due  to  uniformly  distributed  loads,  the  follow- 
ing rules  are  recommended: 

(a)  That  for  floor  slabs  the  bending  moments  at  center  and  at 

w  I2 

support  be  taken  at  -r^  for  both  dead  and  live  loads,  where  w 
*.& 

represents  the  load  per  linear  foot  and  I  the  span  length. 

(6)  That  for  beams  the  bending  moment  at  center  and  at 

support  for  interior  spans  be  taken  at  ^0^  and  f°r  end  spans 

La 

wP 
it  be  taken  at  -r^r  for  center  and  adjoining  support  for  both  dead 

and  live  loads. 

In  the  case  of  beams  and  slabs  continuous  for  two  spans  only, 
or  of  spans  of  unusual  length,  more  exact  calculations  should  be 
made.  Special  consideration  is  also  required  in  the  case  of  con- 
centrated loads. 

Where  beams  are  reinforced  on  the  compression  side,  the  steel 
may  be  assumed  to  carry  its  proportion  of  stress  in  accordance 
with  the  provisions  of  Chap.  VII,  Section  3,  c,  paragraph  6.  In 
the  case  of  continuous  beams,  tensile  and  compressive  reinforce- 
ment over  supports  must  extend  sufficiently  beyond  the  support 
to  develop  the  requisite  bond  strength. 

7.  Bond    Strength    and    Spacing  of    Bars.     Adequate    bond 

I  =  length,  and  b  =  breadth  of  slab.  For  various  ratios  of  l/b  the  values  of  r 
are  as  follows: 

l/b  r 

1  0.50 

1.1  0.59 

1.2  0.67 

1.3  0.75 

1.4  0.80 

1.5  0.83 

Using  the  values  above  specified,  each  set  of  reinforcement  is  to  be  calculated 
in  the  same  manner  as  slabs  having  supports  on  two  sides  only,  but  the  total 
amount  of  reinforcement  thus  determined  may  be  reduced  25  per  cent,  by 
gradually  increasing  the  rod  spacing  from  the  third  point  to  the  edge  of  the 
slab. 


150  REINFORCED  CONCRETE 

strength  should  be  provided  in  accordance  with  the  formula 
hereinafter  given.  Where  a  portion  of  the  bars  is  bent  up  near 
the  end  of  a  beam,  the  bond  stress  in  the  remaining  straight  bars 
will  be  less  than  is  represented  by  the  theoretical  formula. 

Where  high  bond  resistance  is  required,  the  deformed  bar  is  a 
suitable  means  of  supplying  the  necessary  strength.  Adequate 
bond  strength  throughout  the  length  of  a  bar  is  preferable  to  end 
anchorage,  but  such  anchorage  may  properly  be  used  in  special 
cases.  Anchorage  furnished  by  short  bends  at  a  right  angle 
is  less  effective  than  hooks  consisting  of  turns  through  180°. 

The  lateral  spacing  of  parallel  bars  should  not  be  loss  than  two 
and  one-half  diameters,  center  to  center,  nor  should  the  distance 
from  the  side  of  the  beam  to  the  center  of  the  nearest  bar  be  less 
than  two  diameters.  The  clear  spacing  between  two  layers  of 
bars  should  not  be  less  than  J-in. 

8.  Shear  and  Diagonal  Tension.  Calculations  for  web 
resistance  shall  be  made  on  the  basis  of  maximum  shearing  stress 
as  determined  by  the  formulas  hereinafter  given.  When  the 
maximum  shearing  stresses  exceed  the  value  allowed  for  the 
concrete  alone,  web  reinforcement  must  be  provided  to  aid  in 
carrying  the  diagonal  tension  stresses.  This  web  reinforcement 
may  consist  of  bent  bars,  or  inclined  or  vertical  members  at- 
tached to  or  looped  about  the  horizontal  reinforcement.  Where 
inclined  members  are  used,  the  connection  to  the  horizontal 
reinforcement  shall  be  such  as  to  insure  against  slip. 

Experiments  bearing  on  the  design  of  details  of  web  reinforce- 
ment are  not  yet  complete  enough  to  allow  more  than  general 
and  tentative  recommendations  to  be  made.  It  is  well  estab- 
lished, however,  that  a  very  moderate  amount  of  reinforcement, 
such  as  is  furnished  by  a  few  bars  bent  up  at  small  inclination, 
increases  the  strength  of  a  beam  against  failure  by  diagonal 
tension  to  a  considerable  degree;  and  that  a  sufficient  amount 
of  web  reinforcement  can  readily  be  provided  to  increase  the 
shearing  resistance  to  a  value  from  three  or  more  times  that 
found  when  the  bars  are  all  horizontal  and  no  web  reinforcement 
is  used.  The  following  allowable  values  for  the  maximum  shear- 
ing stress  are  therefore  recommended,  based  on  the  working 
stresses,  5  on  page  154. 

(a)  For  beams  with  horizontal  bars  only  40  Ib.  per  sq.  in. 


ELEMENTARY  DESIGN  151 

(6)  For  beams  in  which  a  part  of  the  horizontal  reinforcement 
is  used  in  the  form  of  bent-up  bars,  arranged  with  due  respect 
to  the  shearing  stresses,  a  higher  value  may  be  allowed,  but  not 
exceed  60  Ib.  per  sq.  in. 

(c)  For  beams  thoroughly  reinforced  for  shear  a  value  not 
exceeding  120  Ib.  per  sq.  in. 

In  the  calculation  of  web  reinforcement  to  provide  the  strength 
required  under  c  above,  the  concrete  may  be  counted  upon  as 
carrying  one-third  of  the  shear.  The  remainder  is  to  be  provided 
for  by  means  of  metal  reinforcement  consisting  of  bent  bars  or 
stirrups,  but  preferably  both.  The  requisite  amount  of  such 
reinforcement  may  be  estimated  on  the  assumption  that  the 
entire  shear  on  a  section,  less  the  amount  assumed  to  be  carried 
by  the  concrete,  is  carried  by  the  reinforcement  in  a  length  of 
beam  equal  to  its  depth. 

The  longitudinal  spacing  of  stirrups  or  bent  rods  shall  not 
exceed  three-fourths  the  depth  of  the  beam. 

It  is  important  that  adequate  bond  strength  be  provided  to 
develop  fully  the  assumed  strength  of  all  shear  reinforcement. 

Inasmuch  as  small  deformations  in  the  horizontal  reinforce- 
ment tend  to  prevent  the  formation  of  diagonal  cracks,  a  beam 
will  be  strengthened  against  diagonal  tension  failure  by  so 
arranging  the  horizontal  reinforcement  that  the  unit  stresses  at 
points  of  large  shear  shall  be  relatively  low. 

9.  Columns.  It  is  recommended  that  the  ratio  of  unsupported 
length  of  column  to  its  least  width  be  limited  to  15. 

The  effective  area  of  the  column  shall  be  taken  as  the  area 
within  the  protective  covering,  as  denned  under  Fireproofing 
or  in  the  case  of  hooped  columns  or  columns  reinforced  with 
structural  shapes  it  shall  be  taken  as  the  area  within  the  hooping 
or  structural  shapes. 

Columns  may  be  reinforced  by  means  of  longitudinal  bars,  by 
bands  or  hoops,  by  bands  or  hoops  together  with  longitudinal 
bars,  or  by  structural  forms  which  in  themselves  are  sufficiently 
rigid  to  act  as  columns.  -The  general  effect  of  bands  or  hoops 
is  greatly  to  increase  the  "  toughness  "  of  the  column  and  its 
ultimate  strength,  but  hooping  has  little  effect  upon  its  behavior 
within  the  limit  of  elasticity.  It  thus  renders  the  concrete  a 
safer  and  more  reliable  material  and  should  permit  the  use  of  a 


152  REINFORCED  CONCRETE 

somewhat  higher  working  stress.  The  beneficial  effects  of 
" toughening"  are  adequately  provided  by  a  moderate  amount 
of  hooping,  a  larger  amount  serving  mainly  to  increase  the 
ultimate  strength  and  the  possible  deformation  before  ultimate 
failure. 

The  following  recommendations  are  made  for  the  relative 
working  stresses  in  the  concrete  for  the  several  types  of  columns : 

(a)  Columns  with  longitudinal  reinforcement  only,  the  unit 
stress  recommended  for  axial  compression  under  Working 
Stresses,  Section  3. 

(6)  Columns  with  reinforcement  of  bands  or  hoops,  as  here- 
inafter specified,  stresses  20  per  cent,  higher  than  given  for  a. 

(c)  Columns  reinforced  with  not  less  than  1  per  cent,  and  not 
more  than  4  per  cent,  of  longitudinal  bars  and  with  bands  or 
hoops,  stresses  45  per  cent,  higher  than  given  for  a. 

(d)  Columns   reinforced   with   structural   steel   column   units 
which  thoroughly  encase  the  concrete  core,  stresses  45  per  cent, 
higher  than  given  for  a. 

In  all  cases,  longitudinal  steel  is  assumed  to  carry  its  propor- 
tion of  stress  in  accordance  with  Section  3.  The  hoops  or  bands 
are  not  to  be  counted  upon  directly  as  adding  to  the  strength 
of  the  column. 

Bars  composing  longitudinal  reinforcement  shall  be  straight, 
and  shall  have  sufficient  lateral  support  to  be  securely  held  in 
place  until  the  concrete  has  set. 

Where  bands  or  hoops  are  used,  the  total  amount  of  such 
reinforcement  shall  not  be  less  than  1  per  cent,  of  the  volume  of 
the  column  enclosed.  The  clear  spacing  of  such  bands  or  hoops 
shall  not  be  greater  than  one-fourth  the  diameter  of  the  enclosed 
column.  Adequate  means  must  be  provided  to  hold  bands  or 
hoops  in  place  so  as  to  form  a  column,  the  core  of  which  shall 
be  straight  and  well  centered. 

Bending  stresses  due  to  eccentric  loads  must  be  provided  for 
by  increasing  the  section  until  the  maximum  stress  does  not 
exceed  the  values  above  specified. 

10.  Reinforcing  for  Shrinkage  and  Temperature  Stresses. 
Where  large  areas  of  concrete  are  exposed  to  atmospheric  con- 
ditions, the  changes  of  form  due  to  shrinkage  (resulting  from 


ELEMENTARY  DESIGN  153 

hardening)  and  to  action  of  temperature  are  such  that  large 
cracks  will  occur  in  the  mass,  unless  precautions  are  taken  to  so 
distribute  the  stresses  as  either  to  prevent  the  cracks  altogether 
or  to  render  them  very  small.  The  size  of  the  cracks  will  be 
directly  proportional  to  the  diameter  of  the  reinforcing  bars 
and  inversely  proportional  to  the  percentage  of  reinforcement 
and  also  to  its  bond  resistance  per  unit  of  surface  area.  To  be 
most  effective,  therefore,  reinforcement  should  be  placed  near 
the  surface  and  well  distributed,  and  a  form  of  reinforcement 
used  which  will  develop  a  high  bond  resistance. 

WORKING  STRESSES 

1.  General  Assumptions.  The  following  working  stresses  are 
recommended  for  static  loads.  Proper  allowances  for  vibration 
and  impact  are  to  be  added  to  live  loads  where  necessary  to  pro- 
duce an  equivalent  static  load  before  applying  the  unit  stresses 
in  proportioning  parts. 

In  selecting  the  permissible  working  stress  to  be  allowed  on 
concrete,  we  should  be  guided  by  the  working  stresses  usually 
allowed  for  other  materials  of  construction,  so  that  all  structures 
of  the  same  class  but  composed  of  different  materials  may  have 
approximately  the  same  degree  of  safety. 

The  stresses  for  concrete  are  proposed  for  concrete  composed 
'of  one  part  Portland  cement  and  six  parts  of  aggregate,  capable 
of  developing  an  average  compressive  strength  of  2000  pounds 
per  square  inch  at  28  days,  when  tested  in  cylinders  8  inches  in 
diameter  and  16  inches  long,  under  laboratory  conditions  of 
manufacture  and  storage,  using  the  same  consistency  as  is  used 
in  the  field.  In  considering  the  factors  recommended  with 
relation  to  this  strength,  it  is  to  be  borne  in  mind  that  the 
strength  at  28  days  is  by  no  means  the  ultimate  which  will  be 
developed  at  a  longer  period,  and  therefore  they  do  not  corre- 
spond with  the  real  factor  of  safety.  On  concretes,  in  which  the 
material  of  the  aggregate  is  inferior,  all  stresses  should  be  pro- 
portionally reduced,  and  similar  reduction  should  be  made  when 
leaner  mixes  are  to  be  used.  On  the  other  hand,  if,  with  the  best 
quality  of  aggregates,  the  richness  is  increased,  an  increase  may 
be  made  in  all  working  stresses  proportional  to  the  increase  in 
compressive  strength  at  28  days,  but  this  increase  shall  not 
exceed  25  per  cent. 


154  REINFORCED  CONCRETE 

2.  Bearing.     When  compression  is  applied  to  a  surface  of 
concrete  larger  than  the  loaded  area,  a  stress  of  32.5  per  cent, 
of  the  compressive  strength  at  28  days,  or  650  pounds  per  square 
inch  on  the  above-described  concrete,  may  be  allowed.     This 
pressure  is  probably  unnecessarily  low  when  the  ratio  of  the 
stressed  area  to  the  whole  area  of  the  concrete  is  much  below 
unity,  but  is  recommended  for  general  use  rather  than  a  variable 
unit  based  upon  this  ratio. 

3.  Axial    Compression.     For    concentric    compression    on    a 
plain  concrete  column  or  pier,  the  length  of  which  does  not 
exceed  12  diameters,.  22.5  per  cent,  of  the  compressive  strength 
at  28  days,  or  450  Ib.  per  sq.  in.  on  2000  Ib.  concrete,  may  be 
allowed. 

For  other  forms  of  columns  the  stresses  obtained  from  the 
ratio  given  under  Design,  Section  9,  may  govern. 

4.  Compression  in  Extreme  Fiber.     The  extreme  fiber  stress 
of  a  beam,  calculated  on  the  assumption  of  a  constant  modulus 
of  elasticity  for  concrete  under  working  stresses,  may  be  allowed 
to  reach  32.5  per  cent,  of  the  compressive  strength  at  28  days, 
or  650  Ib.  per  sq.  in.  for  2000-lb.  concrete.     Adjacent  to  the  sup- 
port of  continuous  beams,  stresses  15  per  cent,  higher  may  be 
used. 

5.  Shear  and  Diagonal  Tension.     Where  pure  shearing  stress 
occurs,   that  is,   uncombined  with  compression  normal  to  the 
shearing  surface,  and  with  all  tension  normal  to  the  shearing 
plane  provided  for  reinforcement,  a  shearing  stress  of  6  per  cent, 
of  the  compressive  strength  at  28  days,  or  120  Ib.  per  sq.  in. 
on  2000-lb.  concrete,  may  be  allowed.     Where  the  shear  is  com- 
bined with  an  equal  compression,  as  on  a  section  of  a  column  at 
45°  with  the  axis,  the  stress  may  equal  one-half  the  compressive 
stress  allowed.     For  ratios  of  compressive  stress  to  shear  inter- 
mediate between  0  and  1,  proportionate  shearing  stresses  shall 
be  used. 

In  calculations  on  beams  in  which  diagonal  tension  is  con- 
sidered to  be  taken  by  the  concrete,  the  vertical  shearing  stresses 
should  not  exceed  2  per  cent,  of  the  compressive  strength  at 
28  days,  or  40  Ib.  per  sq.  in.  for  2000-lb.  concrete. 

6.  Bond.     The  bonding  stress  between  concrete  and  plain 
reinforcing  bars  may  be  assumed  at  4  per  cent,  of  the  compressive 
strength  at  28  days,  or  80  Ib.  per  sq.  in.  for  2000-lb.  concrete; 


ELEMENTARY  DESIGN  155 

i-*9 

in  the  case  of  drawn  wire,  2  per  cent,  or  40  Ib.  on  2000-lb.  con- 
crete. 

7.  Reinforcement.     The  tensile  stress  in  steel  should  not  ex- 
ceed 16,000  Ib.  per  sq.  in.     The  compressive  stress  in  reinforcing 
steel  should  not  exceed  16,000  Ib.  per  sq.  in.,  or  15  times  the 
working  compressive  stress  in  the  concrete. 

In  structural  steel  members,  the  working  stresses  adopted  by 
the  American  Railway  Engineering  and  Maintenance  of  Way 
Association  are  recommended. 

8.  Modulus  of  Elasticity.     The  value  of  the  modulus  of  elas- 
ticity of  concrete  has  a  wide  range,  depending  upon  the  materials 
used,  the  age,  the  range  of  stresses  between  which  it  is  considered, 
as  well  as  other  conditions.     It  is  recommended  that  in  all  com- 
putations it  be  assumed  as  one-fifteenth  that  of  steel,  as,  while 
not  rigorously  accurate,  this  assumption  will  give  safe  results. 
(See  table  on  next  page.) 

Design  of  a  T-beam 

It  is  required  to  design  a  beam  of  this  description  to  carry  a 
load  of  2000  pounds,  besides  its  own  weight,  per  linear  foot  of 
span,  which  is  36  feet.  If  the  depth  and  width  of  neither  the 
web  nor  the  flange  be  limited  by  special  conditions,  there  are, 
with  the  amount  of  reinforcement,  five  quantities  to  be  assumed 
or  found.  These  quantities  are:  6,  &',  i,  d,  and  p.  Often  the 
flange  is  a  part  of  a  floor  and  must  be  designed  as  a  slab  sup- 
ported by  two  adjacent  beams  and  so  i  is  thus  fixed.  The 
part'  of  the  slab  that  may  be  considered  as  a  part  of  the 
T-beam  is  indefinite,  but  practically  the  width  is  limited  as 
described  on  page  148.  The  width  of  the  web,  &',  is  limited  by 
the  condition  that  the  steel  must  be  properly  spaced'  and  also 
that  the  strength  of  b'd  in  shear  shall  be  sufficient.  The  depth 
is  limited  practically  by  the  requirements  as  to  head  room  under- 
neath, while  p  may  be  assumed  within  rather  narrow  limits.  In 
this  problem  it  is  assumed  that  the  beam  is  isolated  and  sup- 
ported at  the  ends. 

A  concrete  of  medium  quality  will  be  assumed  and  working 
values  of  concrete  in  compression,  in  shear,  and  of  steel  in  tension 
are  taken  as  600,  30,  and  16,000  Ib.  per  sq.  in.  respectively.  In 
sections  containing  stirrups  and  bent-up  bars  the  shear  is  in- 
creased to  90  Ib.  per  sq.  in. 


156 


REINFORCED  CONCRETE 


AREAS,  WEIGHTS,  AND  SPACING  OF  ROUND  RODS 


Diam- 
eter, 
Inches 

Area 
Square 
Inches 

Perimeter, 
Inches 

Weight 
per  foot 
Pounds 

For  Depth  of  10  Inches  the  Spacing,  in 
Inches,  is  for  Steel  to  the  amount  of 

0.2% 

0.4% 

0.6% 

0.8% 

1.0% 

1 

4 

.0491 

.7854 

.167 

2.45 

1.23 

0.82 

0.61 

0.491 

A 

.0767 

.9818 

.261 

3.83 

1.92 

1.28 

0.96 

0.767 

1 

.1104 

1.1781 

.376 

5.52 

2.76 

1.84 

1.38 

1.104 

A 

.1503 

1.3745 

.511 

7.51 

3.76 

2.50 

1.88 

1.503 

i 

.1963 

1.5708 

.668 

9.82 

4.91 

3.27 

2.45 

1.963 

A 

.2485 

1.7672 

.845 

12.42 

6.21 

4.14 

3.10 

2.485 

1 

.3068 

1.9635 

1.043 

15.34 

7.67 

5.11 

3.83 

3.068 

H 

.3712 

2.1599 

1.262 

18.57 

9.28 

6.18 

4.64 

3.712 

t 

.4418 

2.3562 

1.502 

22.09 

11.04 

7.36 

5.52 

4.418 

it 

.5185 

2.5526 

1.763 

25.93 

12.96 

8.64 

6.48 

5.185 

1 

.6013 

2.7489 

2.044 

30.06 

15.03 

10.02 

7.51 

6.013 

it 

.6903 

2.9453 

2.347 

34.53 

17.26 

11.51 

8.63 

6.903 

1 

.7854 

3.1416 

2.670 

39.27 

19.64 

13.09 

9.82 

7.854 

1* 

.9940 

3.5343 

3.380 

49.70 

24.85 

16.57 

12.42 

9.940 

U 

1.2272 

3.9270 

4.172 

61.36 

30.68 

20.45 

15.34 

12.272 

1! 

1.4849 

4.3197 

5.049 

74.25 

37.12 

24.75 

18.56 

14.849 

H 

1.7671 

4.7124 

6.008 

88.35 

44.17 

29.45 

22.09 

17.671 

AREAS,  WEIGHTS,  AND  SPACING  OF  SQUARE  BARS 


1 

.0625 

1.00 

.212 

3.13 

1.56 

1.04 

0.78 

0.625 

A 

.0977 

1.25 

.332 

4.88 

2.44 

1.63 

1.22 

0.977 

! 

.1406 

1.50 

.478 

7.03 

3.56 

2.34 

1.76 

1.406 

A 

.1914 

1.75 

.651 

9.57 

4.78 

3.19 

2.39 

1.914 

i 

.2500 

2.00 

.850 

12.50 

6.25 

4.16 

3.13 

2.500 

A 

.3164 

2.25 

1.076 

15.82 

7.91 

5.27 

3.96 

3.164 

1 

.3906 

2.50 

1.328 

19.53 

9.76 

6.51 

4.88 

3.906 

H 

.4727 

2.75 

1.607 

23.63 

11.81 

7.88 

5.91 

4.727 

i 

.5625 

3.00 

1.913 

28.12 

14.06 

9.37 

7.03 

5.625 

it 

.6602 

3.25 

2.245 

33.01 

16.50 

11.00 

8.25 

6.602 

1 

.7656 

3.50 

2.603 

38.28 

19.14 

12.76 

9.57 

7.656 

it 

.8789 

3.75 

2.988 

43.94 

21.97 

14.65 

10.98 

8.789 

i 

1.0000 

4.00 

3.400 

50.00 

25.00 

16.66 

12.50 

10.000 

it 

1.2656 

4.50 

4.303 

63.28 

31.64 

21.09 

15.82 

12.656 

U 

1.5625 

5.00 

5.313 

78.12 

39.06 

26.04 

19.53 

15.625 

H 

1.8906 

5.50 

6.428 

94.53 

47.26 

31.51 

23.63 

18.906 

1* 

2.2500 

6.00 

7.650 

112.50 

56.25 

37.50 

28.12 

22.500 

ELEMENTARY  DESIGN  157 

Cross-section  According  to  Shear.  The  first  requirement  in 
making  a  design  is  to  know  the  weight  of  the  structure  or,  as 
this  is  always  unknown,  to  assume  a  weight  such  that  the  pre- 
liminary design  may  not  have  to  be  much  changed  when  the 
true  weight  becomes  known.  It  is  customary  to  assume  the 
dead  load  as  |  or  \  the  live  load  and,  in  this  case,  the  former 
fraction  will  be  used.  With  this  assumption  the  cross-section 
at  the  end  necessary  to  carry  the  vertical  shear  may  be  found. 
The  dead  load  is  J  X  2000  Ib.  per  linear  foot  and  the  shear  is 
£  X  2000  X  18  =  48,000  Ib.  The  section,  b'd,  is  then  48,000 
-f.  90  =  533  sq.  in.  since  all  the  shear  is  to  be  taken  by  the  web. 
The  web  must  be  wide  enough  to  give  proper  spacing  for  the 
reinforcing  bars  and  is  comparatively  larger  in  small  beams 
than  in  large  ones,  as  the  latter  frequently  have  the  steel  in  two 
horizontal  rows  and  as  the  outside  covering  of  concrete  is  con- 
stant. Usually,  if  d  be  taken  as  2  or  2J  times  b'  in  small  beams 
and  as  3  to  4  times  d  in  large  ones,  it  will  be  found  that  but 
little  change  in  b'  will  be  necessary  when  the  computations  are 
remade  with  known  weights.  If  d  =  36'  in  this  case,  b'  becomes 
\/533  -I-  33  =  13.5  and  d  is  40.5  inches. 

Flange  Width  According  to  Moments.  For  the  assumed 
weights,  the  moment  is  M  =  J  X  2670  X  36  X  36  X  12  = 
5,200,000  Ib.  in.  Now,  in  Plate  III,  page  51,  either  t,  b,  or  p 
may  be  assumed  and,  as  t  is  not  limited,  p  will  be  taken  as  0.5  per 
cent,  and  t,  usually  between  0.5  b'  and  0.75  b',  as  10  inches.  Then 
N3  =  0.139  and  b  =  5,200,000  ^  (.139  X  40.5  X  40.5  X  60)  = 
38  inches.  The  dimensions  have  now  been  determined  tenta- 
tively and  the  weight  may  be  recomputed.  The  area  of  cross- 
section  is  38  X  10  +  b'(h — 10)  in  which  h  may  be  taken,  for  trial, 
as  44  inches.  The  weight  of  the  beam  is  (38  X  10  +  34  X  13.5) 
150  -5-  144  =  875  Ib.  per  linear  foot,  and  the  whole  load  is  2875 
Ib.  per  linear  foot,  whereas  2670  was  assumed.  The  bending 
moment,  hence,  will  be  increased  7.7  per  cent,  and  p  or  one  of  the 
dimensions  must  be  changed  accordingly.  The  depth  may  be 
made  42  inches,  b  changed  to  41.5  inches  or  p  to  0.7  per  cent. 
In  this  case  the  change  will  be  made  in  the  breadth.  The  shear 
is  also  somewhat  increased  and  d  may  have  to  be  changed  on 
this  account. 

The  Reinforcement.  The  area  of  reinforcement  is  41.5  X 
40.5  X  0.005  =  8.4  sq.  in.  This  amount  is  supplied  by  seven 


158 


REINFORCED  CONCRETE 


bars  1|  inches  square  having  an  area  of  7  X  1.265  =  -8.86  sq.  in. 
The  steel  will  be  arranged  in  two  horizontal  rows  of  three  above 
and  four  below.  The  necessary  width  is  from  page  82,  b  =  1.125 
(1.5  +  2.5  X  4)  =  12.94  inches  and  so  the  assumed  width  is 
sufficient.  The  center  of  gravity  of  the  bars  is  4  of  the  vertical 
distance  between  the  lines  through  the  centers  of  the  bars  in  the 
two  rows.  The  clear  distance  between  the  two  rows  must  be 
half  an  inch,  so  the  whole  depth  of  the  beam  is  40.5  +  0.64  + 
0.5  +  2  =  43.64  or  44  inches  as  assumed.  The  depth  at  the 
end  of  the  beam,  to  the  middle  of  the  lower  bars,  is  41  f  in.  .  The 
shearing  stress  is  51,750  +  (41.4  X  13.5)  =  91.2  Ib.  sq.  in.  So 
b'  is  made  14  in. 

Points  for  Bending  up  Bars.     The  first  bars  to  be  bent  up  are 
those  in  the  upper  row  nearest  the  middle  of  &';  in  this  problem 


Taken  by 
Bent-up  Rods  and 
Stirrups 


FIG.  59. 

one  bar  will  be  bent  up,  then  others  in  pairs.  In  Fig.  38  it  is 
seen  that  \  of  the  bars  may  be  bent  up  at  0.1 9Z,  f  at  0.34,  and 
|  at  0.42  I  or  at  6.9,  12.3,  and  15.2  feet  from  the  middle  of  the 
beam.  As  was  stated  on  page  77,  the  spacing  of  the  web  rein- 
forcement depends  upon  the  strength  of  the  bar  and  of  the  bond 


ELEMENTARY  DESIGN 


159 


and  also  upon  the  depth  of  the  beam.  As  these  bars  are,  as  is 
usual,  to  be  bent  to  an  angle  of  45°  with  the  horizontal,  they  can- 
not be  spaced  more  than  40.5  inches  apart,  and  the  first  bar 
must  be  continued  straight  beyond  the  length  required  by  the 
bending  moment. 

Spacing  the  Web  Reinforcement.  The  vertical  shear  at  the 
end  of  the  beam  is  18  X  2875  =  51,750  pounds  and  at  the  middle 
v  =  2000  X  18  '-T-  4  =  9000  pounds.  The  value  of  j  is,  by  (24) 
page  49,  j  =  d  -  [3  X  40.5  X  15  -  20  (15  +  16.7)]  *  -s-  [6  X 
40.5  X  15  -  30  (15  X  16.7)]  =  36.3  inches.  The  concrete  is 
capable  of  carrying  30  X  36.3  X  14  =  15,200  pounds  of  shear 
while  the  web  reinforcement  takes  the  remainder  or  36,600 
pounds.  The  diagram  of  shears  is  shown  in  Fig.  59.  The 
bent-up  rods  nearest  the  ends  should  pass  over  the  support 
well  below  the  neutral  axis  and  so  the  point  of  bending  should 
not  be  more  than  J  d  from  the  support.  The  shear  to  be  carried 
by  these  two  rods  may,  then,  be  taken  as  about  37,000  pounds. 
From  (62)  s  =  2  X  1.265  X  16,000  X  36.3  -r-  (37,000  X  .707)  = 
55  inches,  so  the  bars  will  be  spaced  40  inches  apart.  At  80 
inches  from  the  support  the  shear  is  21,000  pounds  and  s  =  1.265 
X  16,000  X  36.3  -^  (21,000  X  .707)  =  49  inches.  As  both 
values  of  s  are  greater  than  the  depth  the  spacing  will  be  40  inches, 
as  shown,  the  first  bend  being  not  over  %d  from  the  support. 

Bond  Stresses  of  the  Bars.  As  has  been  shown  before,  the 
grip  of  the  bent-up  rods  should 
be  about  50  diameters,  which 
distance  is  secured  by  bend- 
ing them  in  the  upper  part  of 
the  beam  as  shown  in  Fig. 
59.  The  rods  nearest  the  end 
of  the  beam  may  be  bent  as  are 
the  others  if  the  beam  extends 
far  enough  over  the  support. 

The  Stirrups.  In  Fig.  59  it 
is  seen  that  a  part  of  the  shear 
is  provided  for  by  neither  the 
concrete  nor  by  the  bent-up 
rods  and  so  stirrups  are  to  be  inserted.  To  secure  better  bond 
small  rods  are  used  and  a  diameter  of  f  inch  will  be  selected. 
Where  the  first  stirrup  is  to  be  placed  the  shear  is  about  18,000 


}<— 13.5-* 

FIG.  60. 


160 


REINFORCED  CONCRETE 


pounds.  The  stirrups  will  be  made  in  two  continuous  loops,  pass- 
ing under  the  horizontal  rods  as  shown  in  Fig.  60.  Then  the  area 
is  4  X  0.11  =  0.44  square  inches.  Substituting  in  (59)  s  =  0.44 
X  16,000  X  36.3  -T-  18,000  =  14.2  inches.  Since  the  shear  to 
be  carried  by  the  stirrups  decreases  uniformly  over  about  80 
inches,  the  average  spacing  will  be  14.2  X  2  =  28.4  inches,  and 
three  spaces  and  four  stirrups  will  suffice.  The  triangle  in  the 
shear  diagram,  representing  the  shear  taken  by  the  stirrups,  is 
to  be  divided  into  three  equal  areas  or,  in  general,  into  n  —  1 
trapezoids  and  one  triangle  having  equal  areas,  n  being  the 
number  of  spaces.  If  I'  be  the  distance  ab,  the  horizontal  dimen- 
sions of  these  figures  will  be 

..  \fn  —  \/n  — 

/- 
\/n 


lf 


7/  \/n  —  1  —  \/n  —  2 


, 


_ 

\/n 


the  first  expression  being,  practically,  the  same  as  s  computed 
from  (59),  and  the  last  one  the  base  of  the  triangle.  The  numeri- 
cal equivalents  of  the  above  expressions  are  tabulated  in  Fig.  61 
for  values  of  n  not  exceeding  10,  and  beyond  this  number  the 
smaller  divisions  become  nearly  uniform. 

In  this  case  there  are  three  spaces  which  will  be  given  by  mul- 
tiplying I'  by  the  fractions  in  the  column  having  3  at  the  top  or 
80  X  .578  =  46,  80  X  .238  =  19  and  80  X  .184  =  15.  These 
spaces  may  be  laid  off  on  I  and  the  stirrups  are  placed  at  the 
centers  of  gravity  of  the  trapezoids  so  formed,  giving  the  spacing 
shown  in  the  figure. 


NUMBER  OF  SPACES 


1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

- 

1.00 

.293 

.184 

.134 

.105 

.087 

.074 

.064 

.057 

.051 

.707 

.238 

.159 

.120 

.096 

.080 

.070 

.061 

.054 

i. 

.578 

.207 

.142 

.109 

.089 

.075 

.066 

.058 

IB 

.500 

.185 

.131 

.101 

.084 

.071 

.062 

Q 

m 

.448 

.169 

.121 

.096 

.079 

.068 

«B 

.408 

.157 

.112 

.089 

.075 

& 

.378 

.146 

.106 

.085 

.353 

.138 

.101 

Table  Giving  Spacing  of  Stirrups 

.333 

.130 

j 

.316 

FIG.  61. 


ELEMENTARY  DESIGN  161 

It  is  usual  to  insert  stirrups  also  in  the  part  of  the  beam  between 
a  and  the  end  as  additional  security.  This  is  to  be  done  here, 
although  the  figure  shows  only  those  taking  the  shear  not  other- 
wise provided  for.  The  minimum  spacing  is  to  be  continued 
to  the  support. 

If  the  numbers  in  Fig.  61  be  divided  by  the  sine  of  the 
angle  the  bent-up  rods  make  with  the  horizontal,  they  apply  to 
the  spacing  of  such  rods. 

Investigation  for  the  Flange.  Should  it  be  possible  that  the 
load  may  be  applied  to  one  flange  rather  than  uniformly  over  the 
whole  top  it  may  be  necessary  to  provide  transverse  reinforce- 
ment in  the  upper  side  of  the  beam.  If  half  the  live  load  be 
placed  toward  one  side  of  the  top,  and  one  foot  of  length  be  con- 
sidered as  a  cantilever  beam  having  a  length  of  14  inches,  the 
bending  moment  will  be  (1000  +  10  X  14  X  150)  X  J  X  14 
=  8000  Ib.  in.  From  (7)  page  39,  A  is  found  to  be  0.08  square 
inches  per  linear  foot.  If  the  stirrups  be  extended  transversely 
some  distance  beyond  the  width  of  the  web,  ample  security  will 
be  given.  In  case  the  flange  is  a  part  of  a  floor  system  the  rein- 
forcement in  the  lower  side  of  the  slabs  is  bent  up  and  continued 
over  the  beams.  An  angle,  such  as  those  between  the  flange 
and  the  stem,  is  always  a  source  of  weakness  and  it  is  good 
practice  to  bevel,  or  round  off,  the  corners  of  the  forms  to  make 
the  change  in  depth  less  abrupt. 

Diagram  for  Solution  of  Slabs  and  Beams.  For  office  use  it  is 
desirable  to  have  a  quick  and  accurate  method  of  finding  dimen- 
sions of  beams  and  slabs  without  making  use  of  the  ordinary 
formulas.  In  Vol.  LIV,  Part  E,  page  537,  of  Trans.  Am.  Soc. 
Civil  Engineers,  Dr.  F.  von  Emberger  shows  a  diagram  which, 
drawn  to  proper  scale,  is  very  useful  in  this  connection.  This 
diagram  is  shown  to  small  scale  in  Fig.  62.  As  the  article  men- 
tioned above  is  readily  accessible  to  very  many  no  description 
of  the  method  of  construction  will  be  given  here. 

Several  examples  are  given  showing  the  use  of  the  diagram 
which  is  readily  constructed  from  the  given  equations. 


Sequence  of  Computations  in  the  Design  of  T-beams.  To  the 
end  that  such  work  shall  proceed  in  logical  order,  the  steps,  as 
given  above,  will  be  repeated  in  the  form  of  instructions. 


162 


REINFORCED  CONCRETE 


General  Instructions.  All  computations  should  be  in  a  note- 
book, one  side  of  the  leaves  being  reserved  for  miscellaneous  and 
trial  operations,  while  finished  results  are  on  the  opposite  page. 


FIG.  62. 

The  size  of  the  trimmed  sheet  of  paper  shall  be  - 

-  in. 
The  dimensions  of  the  border  line  shall  be  —    -  in.  X  - 

The  scale  to  be  used  shall  be in.  =  1  foot. 

The  plan  shall  be  in  the  lower  left  hand  corner. 

The  side  elevation  shall  be  in  the  upper  left  hand  corner. 


in.   X 


m. 


ELEMENTARY  DESIGN  163 

The  end  elevation,  or  cross-section,  shall  be  in  the  upper  right 
hand  corner. 

The  title,  bill  of  material,  and  notes  shall  be  in  the  lower  right 
hand  corner. 

The  shear  and  moment  diagrams  shall  be  below  the  side  eleva- 
tion. 

Data.  Isolated  T-beam,  supported  at  the  ends.  Length  of 
span  -  -  feet.  Weight  of  concrete,  150  pounds  per  cubic  foot. 
Live  load  -  -  pounds  per  linear  foot,  C  =  500  to  650,  St  = 
15,000  to  16,000,  and  v  =  30  to  40  pounds  per  square  inch.  In 
sections  through  reinforcement  v  =  90  to  120  pounds  per  square 
inch.  Bond  stress  80  pounds  per  square  inch,  n  =  15. 

The  Order  of  Computation.  The  calculations  are  taken  up 
as  follows: 

1.  Assume  the  dead  load  to  be  J  or  \  of  live  load. 

2.  Find  the  maximum  shear  at  supports. 

3.  Compute  the  web  section,  b'd,  to  support  the  shear. 

4.  Assume  relative  proportions  for  6'  and  d.     The  depth  is 
usually  from  26'  to  2|6'  for  small  beams  and  from  36'  to  46'  for 
large  ones. 

5.  Choose  thickness  of  the  flange,  usually  from  |6'  to  f 6'. 

6.  Compute  the  bending  moment  for  assumed  dimensions, 
M  =  IwP. 

7.  Assume  6  or  p  and  find  p  or  6  from  Plate  III,  page  51. 

8.  Compute  the  width  of  web  from  the  spacing  of  the  re- 
inforcement and  change  the  previous  assumption  if  necessary. 
Allow  2  inches  of  concrete  outside  of  rods  in  beams  of  more  than 
20  inches  depth,  and  use  spacing  according  to  page  82.     Find 
center  of  gravity  of  reinforcement. 

9.  Find  total  depth,  h,  and  compute  the  dead  load.     With 
the  new  loads,  recompute  the  shear  and  bending  moment  and 
change  assumed  dimensions  accordingly. 

10.  Determine  the  points  at  which  rods  may  be  bent  up,  using 
Fig.   38.     Provide  the  straight  rods  with   sufficient  length  of 
grip,  making  hooks  at  the  ends  if  necessary.     Use  (66). 

1 1 .  Find  j  and  k  from  (24)  and  (26) . 

12.  Construct  the  shear  diagram  and  determine  the  parts  of 
the  shear  to  be  carried  by  the  concrete,  the  bent-up  rods,  and  the 
stirrups.     The  rods  are  to  be  bent  up  in  pairs.     Those  in  the 
upper  row  and  near  the  middle  of  6'  are  bent  up  first. 


164  REINFORCED  CONCRETE 

13.  Determine    the    spacing   of   the   bent-up   rods    and   the 
stirrups,  using  (62)  and  Fig.  61.     The  spacing  of  rods  and  stirrups 
must  not  exceed  d  and  f  d  respectively. 

14.  Investigate  the  flange  transverse  bending  stresses  under 
eccentric  loading  and  for  shear  in  the  plane  of  the  side  of  the  web 
produced. 

15.  Design  the  support  so  that  the  vertical  compressive  stress 
shall  not  be  excessive,  and  that  sufficient  anchorage  for  reinforce- 
ment be  secured.     Usually  the  support  should  be,  at  least,  bf 
X  id. 

PROBLEMS 

68.  Prove  that  the  spacing  given  in  Fig.  61  may  be  accomplished 
graphically  as  follows:  Let  /'  be  divided  into  n  equal  spaces,  and  from 
each  division  point  let  perpendiculars  be  erected  to  the  semicircle  drawn 
with  V  as  a  diameter.     With  one  end  of  /'  as  a  center,  arcs  are  drawn 
from  the  tops  of  the  perpendiculars  to  I',  thus  dividing  V  as  specified. 

69.  Compute  the  cost  of  the  T-beam  designed  above,  using  concrete 
in  place  at  $8  per  cubic  yard,  and  steel  in  place  at  3  cents  per  pound. 

70.  Find  the  minimum  cost  if  i  be  made  to  vary  between  6  inches  and 
12  inches,  and  p  be  the  other  variable. 

71.  If  t,  b,  and  b'  be  constant,  find  p  and  d  for  minimum  cost. 

Design  of  Reinforced  Retaining  Walls.  The  design  of  retain- 
ing walls  is  not  nearly  as  accurate  and  satisfactory  as  that  of 
dams  and  reservoir  walls.  The  reason  is  that  little  is  known 
regarding  the  lateral  pressure  of  the  earth  filling  except  in  special 
cases.1  Many  valuable  experiments  have  been  made  in  attempts 
to  correct  the  present  uncertainty  regarding  the  action  of  loose 
material  that  is  restrained  to  a  slope  steeper  than  that  of  its 
natural  repose,  but  no  universally  accepted  rules  have  resulted. 
If  materials  described  as  sand,  loam,  shingle,  clay,  and  so  forth, 
were  the  same  everywhere  and  at  all  times,  it  would  be  com- 
paratively easy  to  so  tabulate  the  results  of  observations  and 
experiments  that  formulas  for  pressure,  that  might  be  depended 
upon,  could  be  deduced,  but  such  is  not  the  case.  Even  in  a 
given  bank,  the  pressure  varies  rapidly  and  decidedly  with 
changes  of  conditions  as  to  moisture,  water,  heat,  and  frost. 
In  some  cases  the  initial  pressure  is  the  maximum  while,  on  the 
other  band,  retaining  walls  have  failed  after  decades  of  seeming 
stability. 

1  See  Vol.  LXX.     Transactions  Am.  Soc.  C.  E.,  1910. 


ELEMENTARY  DESIGN  165 

The  lateral  pressure  of  loose  material  undoubtedly  varies 
somewhat  with  the  unit  weight,  but  more  with  the  degree  of 
fluidity,  as  indicated  by  the  angle  of  repose  which  appears  as  a 
factor  in  nearly  all  such  formulas.  As  the  coefficient  of  friction, 
or  the  tangent  of  the  angle  of  repose  is  not  generally  known, 
it  is  customary  to  design  solid  masonry  retaining  walls  according 
to  precedent  rather  than  in  accordance  with  any  formula  for 
earth  pressure.  Reinforced  concrete  retaining  walls  are  of  a 
newer  type  than  the  solid,  and  so  there  are  fewer  examples  of 
designs  that  have  proved  satisfactory.  It  would,  then,  seem 
that  the  proper  method  to  pursue  is  to  design  the  reinforced 
wall  to  have  the  same  degree  of  security  as  has  been  determined 
for  the  solid  type.  This  may  be  done  by  finding  the  pressure 
of  a  backing  against  a  vertical  plane,  of  height  equal  to  that 
of  the  wall,  through  the  back  of  the  footing  which,  combined 
with  the  weight  of  the  wall  and  of  the  prism  of  earth  between 
the  wall  and  the  vertical  plane  mentioned  above,  will  cause  the 
resultant  to  pass  through  the  outside  of  the  middle  third  of 
the  base.  The  pressure  of  the  earth  against  the  vertical  may 
be  taken  as  \  w'h2,  in  which  wf  is  the  unit  weight  of  a  fluid  that 
would  cause  the  same  pressure,  horizontally,  as  does  the  earth 
and  h  is  the  height  of  the  wall.  The  prism  is  supposed  to 
act  vertically,  without  horizontal  pressure  in  addition  to  that 
of  the  earth  back  of  it.  This  method  of  analysis  might  be 
called  that  of  finding  the  equivalent  liquid  pressure  of  earth, 
but  it  is  that  only  to  the  extent  of  considering  the  horizontal 
pressure  as  varying  as  the  square  of  the  height.  The  unit  weight 
of  the  prism  is  the  actual  unit  weight  which  is  very  different 
from  the  liquid  unit  weight  to  be  computed  for  the  earth 
behind  the  prism. 

A  masonry  wall,  with  suitable  foundations,  is  usually  con- 
sidered of  proper  cross-section  if  the  top  has  a  thickness  suffi- 
cient to  withstand  shocks  to  which  it  may  be  liable,  as  two  to 
three  feet,  a  nearly  vertical  face,  as  1  to  12,  and  a  width  of  base 
equal  to  four-tenths  or  five-tenths  of  the  height.1  For  such  a 
wall  it  may  be  readily  shown  that  wf  is  from  18  pounds  to  25 
pounds  per  cubic  foot  when  the  resultant  pressure  passes  through 
the  outside  of  the  middle  third  of  the  base,  and  the  weights  of 

1  See  Bulletin  No.  108,  Am.-  Railway  Engineering  and  Maintenance  of 
Way  Association,  February,  1909. 


166 


REINFORCED  CONCRETE 


masonry  and  of  the  earth  prism  are  150  and  100  pounds  respect- 
ively per  cubic  foot.  With  these  assumptions  the  pressure 
between  the  earth  and  the  masonry  at  the  foundation  is,  at  the 
outer  edge,  two  times  the  average  pressure  and  is  zero  at  the 
inner  edge. 

Reinforced  concrete  retaining  walls  are,  in  general,  of  two 
types,  as  shown  in  Fig.  63.  The  cantilever  type  is  shown  in 
(a),  and  the  counterpart  type  in  (6).  The  former  is  seldom  used 


with  a  height  exceeding  25  feet,  and  the  latter  is  used  for  both 
high  and  low  walls.  The  semi-gravity  type  is  shown  in  (c). 
This  is  used  exclusively  on  some  railroads  in  the  middle  West. 

Design  of  a  Cantilever  Retaining  Wall 

Let  the  height  of  the  wall  above  ground  be  16  feet,  and  let 
the  top  of  the  footing  be  2  feet  below  ground.  The  unit  hori- 
zontal pressure  of  the  earth  backing  is  assumed  to  be  25  pounds 
per  square  foot.  The  safe  compressive  strength  of  the  con- 
crete is  400  pounds  per  square  inch,  the  allowable  tensile  strength 
of  the  steel  is  12,500  pounds  per  square  inch,  the  bond  strength 
is  80  pounds  per  square  inch,  the  diagonal  tensile  strength  of 
the  concrete  is  40  pounds  per  square  inch,  and  n  is  15.  The 
weight  of  concrete  is  150,  and  of  earth  is  100  pounds  per  cubic 
foot. 

The  wall  is  seen  to  be  composed  of  three  cantilevers,  the 
vertical  wall  restraining  the  earth,  the  front  of  the  footing  which 
is  in  tension  on  the  under  side,  and  the  rear  of  the  footing  which 
is  in  tension  on  the  upper  side.  These  will  be  designed  in  order. 

The  Vertical  Wall.  The  pressure  of  the  earth  will  be  possibly 
dependent  upon  the  amount  of  surcharge  above  the  level  of 


ELEMENTARY  DESIGN  167 

the  top  of  the  wall,  so  the  unit  horizontal  earth  pressure  is  as- 
sumed rather  large,  25  pounds  per  square  foot  of  vertical  pro- 
jection of  the  wall.  As  such  a  wall  weighs  less  than  a  solid  wall 
of  the  same  height,  it  may  be  tentatively  assumed  that  the 
base  will  be  0.45  X  20  =  9  feet.  The  bending  moment  of  the 
earth  against  the  vertical  wall  at  the  top  of  the  footing  will  be 
M  =  12.5  X  183  X  12  -r-  3  =  291,600  Ib.  in.  If  p  be  taken  as 
0.005  in  Plate  I,  t  =  V291600  -v-  (.144  X  12.  X  400)  =  20.5 
inches  to  the  middle  of  the  steel,  or  23  inches  over  all.  The 
thickness  required  to  resist  bending  varies  as  h* ,  and  so  decreases 
rapidly  toward  the  top,  but  considerable  thickness  is  required 
at  the  top  to  resist  shocks,  frost,  and  so  forth,  and  this  is  here 
taken  as  10  inches.  The  inside  batter  is  made  uniform  to  reduce 
the  price  of  the  forms,  so  the  thickness  at  6  feet  and  12  feet 
from  the  top  is  14.3  inches  and  18.7  inches  respectively. 

Reinforcement  in  Vertical  Wall.  If  the  reinforcement  be 
made  up  of  f  inch  round  rods,  the  spacing  at  the  footing  will 
be  0.6013  -^  (0.005  X  20.5)  =  5.85  inches.  At  a  point  12  feet 
below  the  top  the  bending  moment  is  87,000  Ib.  in.,  and  at 
6  feet  below  it  is  11,000  Ib.  in.  Referring  to  Plate  I  or  (13), 
it  is  seen  that  less  than  J  and  J  of  the  reinforcement  used  at  the 
footing  will  be  sufficient  at  these  points  respectively.  So  every 
third  rod  is  carried  up  6  feet  and  every  sixth  one  to  the  top  of 
the  wall.  The  most  important  point  to  be  considered  in  the 
design  of  this  type  of  retaining  wall  is  the  anchoring  of  the  rods 
in  the  footing.  This  question  will  be  taken  up  when  the  dimen- 
sions of  the  latter  are  determined. 

It  will  be  noted  that  making  a  uniform  batter  decreases  the 
necessary  amount  of  steel  and  allows  the  use  of  simpler  and 
cheaper  forms.  Longitudinal  rods  are  inserted  to  prevent  large 
cracks.  These  are  more  liable  to  occur  near  the  top  and  on 
the  outside,  and  so  these  parts  are  more  heavily  reinforced  than 
the  others.  If  p  =  0.004  be  inserted  at  the  top,  the  spacing 
of  |-inch  round  rods  will  be  0.196  -f-  (10  X  0.004)  =  5  inches. 
This  average  will  be  preserved  if  the  spacing  be  8  inches  in  front 
and  12  inches  at  the  back  of  the  wall. 

Shear  at  Top  of  Footing.  The  horizontal  pressure  against  a 
foot  of  wall  is  J  X  25  X  18  X  18  =  4050  pounds,  and  the  unit 
shear  is  4050  -^  (12  X  20.5)  =  16.5  Ib.  sq.  in.,  which  may  be 
carried  by  the  concrete. 


168 


REINFORCED  CONCRETE 


Design  of  the  Footing.  Frequently  the  projection  of  the 
footing  in  front  of  the  wall  is  limited  by  property  lines;  other- 
wise, with  a  given  length  of  base,  the  stability  of  the  wall  is 
made  maximum  by  varying  EC  and  DB  in  Fig.  65.  For  sim- 


h- 


w 


A 


F 
FIG.  64. 


C    G      B 

FIG.  65. 


plicity,  let  the  section  be  as  in  Fig.  64,  and  let    moments  be 
taken  about  A.     Then  Wx  +  Ee  +  Fl/2  -  Hh/3  =  M. 

But  e  =  I  +  ^  +  \  and  E  =  w'h'  (l  -  x  -|) 
also  w  and  wr  are  the  unit  weights  of  masonry  and  of  earth, 


=  0 


and,  for  usual  values,  x  =  t  f or  maximum  stability.  It  is  usually 
attempted  to  bring  the  resultant  of  all  forces  inside  the  middle 
third  of  the  base,  and  to  have  this  point  under  the  vertical 
part  of  the  wall.  In  this  problem  the  base  was  assumed  to  be 
9  feet  in  length.  The  projection  in  front  will  be  taken  as  2.5 
feet.  The  height  of  earth  over  D  becomes  21.8  feet.  The 
weights  of  the  prism  of  earth,  the  wall,  and  the  footing  are  10,300, 
3710,  and  2700  pounds  respectively.  These  weights  are  found 
to  act  at  distances  of  3.55,  0.16,  and  1.5  feet  respectively  from 


ELEMENTARY  DESIGN  169 

the  outer  third  point.  The  resisting  moment  about  this  point 
is  then  10,300  X  3.55  +  3710  X  0.16  +  2700  X  1.5  =  41,240 
Ib.  ft.  The  overturning  moment  about  the  same  point  is  12.5  X 
203  -f-  6  =  33,333  Ib.  ft.,  and  the  base  need  not  be  changed  in 
length.  The  weights  act  at  41,240  4-  (10,350  +  3710  +  2700) 
=  2.48  feet  from  the  outer  edge  of  the  middle  third.  The 
pressure  of  the  earth  is  5000  pounds,  and  acts  at  6.67  feet  above 
the  bottom  of  the  footing.  The  resultant  may  be  drawn  to 
scale,  as  shown  in  Fig.  67,  and  the  point  of  application  is  found 
to  be  0.48  feet  to  the  right  of  the  outer  edge  of  the  middle  third 
of  the  base,  and  so  the  assumption  of  9  feet  as  the  length  of 
the  base  was  proper.  Had  the  resultant  fallen  outside  the 
middle  third  a  few  trials  would  have  indicated  the  amount  to 
be  added  to  the  base  at  each  end. 

The  Bearing  Power  of  the  Soil.  Aside  from  the  question  of 
overturning  on  account  of  the  horizontal  pressure  of  the  earth, 
the  footing  must  be  of  such  length  as  to  bring  the  unit  pressure 
within  the  safe  bearing  power  of  the  earth  beneath.  The  unit 
pressure  at  the  outer  edge  of  the  wall  depends  upon  the  weight 
of  the  wall  and  the  prism  of  earth  above  it,  and  also  upon  the 
eccentricity  of  the  point  of  application  of  the  forces  above. 
When  the  resultant  is  at  the  edge  of  the  middle  third  the  maxi- 
mum pressure  on  the  base  is  double  the  mean.  If  the  resultant 
be  outside  the  J  point  and  e  the  eccentricity,  the  maximum 
pressure  is  given  by  2  W+  3  (1/2  —  e),  and  if  inside,  by  W/l  + 
6  We/I2.  In  this  problem  the  maximum  pressure  on  the  soil 
will  not  exceed  16,710  X  2  -f-  9  =  3715  pounds  per  square  foot, 
if  the  resultant  pass  through  the  %  point,  or  3130  pounds  per 
square  foot  if  the  forces  be  as  computed  above. 

The  Outer  Cantilever.  The  length  of  the  cantilever  is  taken 
as  2.5  +  (1.9  -5-  2)  =  3,5  feet.  The  downward  weight  is  2  X  2.5  X 
100  +  2.5  X  2  X  150  =  1250  pounds,  having  a  lever  arm  of  2.2 
feet.  The  upward  pressure  of  the  earth  is  (3150  +  2150)  X  3.5  -=- 
2  =  9300  pounds,  having  a  lever  arm  of  1.86  feet.  The  moment 
about  the  point  under  the  middle  of  the  vertical  slab  is 
9300  X  1.86  -  1250  X  2.2  =  15,350  Ib.  ft.  The  effective  depth 
of  the  footing  may  be  taken  at  24  —  2.5  =  21.5  inches,  as 
assumed,  and  this  is  to  be  investigated.  A  convenient  way  of 
anchoring  the  rods  in  the  vertical  slab  is  to  bend  them  into  the 
horizontal  to  form  the  reinforcement  of  the  outer  cantilever. 


170  REINFORCED  CONCRETE 

If  all  these  rods  be  used  in  this  way  the  value  of  p  is 
0.6013  -f-  (5.85  X  21.5)  -  0.0047.  Using  Plate  I,  15,350  X  12 
=  0.140  X  400  X  12  d2  and  d  is  16.6  inches.  This  shows  that 
not  all  the  vertical  bars  are  needed  for  this  purpose,  but  they 
serve  to  reduce  the  bond  stress.  In  Plate  I,  r  is  33.5  and 
St  =  400  X  33.5  =  13,400  Ib.  sq.  in.  If  the  working  bond  stress 
be  80  Ib.  sq.  in.,  the  length  of  grip  required  is  *"J-80  x  =  13,400 
X  0.6013  and  x  =  37  inches.  This  distance  is  secured  by  turn- 
ing the  rods  upward  as  shown.  The  cantilever  must  be  exam- 
ined as  to  the  bond  stress  developed.  The  formula  is  (2), 
page  25,  and  the  shear  is  found  to  be  7000  pounds,  at  the 
point  C,  for  a  linear  foot  of  wall.  Then  u  =  7000  -r-  (21.5  X  1 
X  2.75  X  12/5.85)  =  66  Ib.  sq.  in.,  which  is  not  in  excess  of  the 
allowable  bond  stress.  Had  this  exceeded  80  pounds  per  square 
inch  the  depth  at  the  edge  of  the  vertical  slab  would  have 
been  increased  accordingly.  The  depth  necessary  to  take  the 
vertical  shear  is  7000  -*-  (12  X  35)  =  16.7  inches,  which  is  also 
less  than  that  assumed.  Half  way  out,  the  shear  is  3730  pounds, 
which  requires  a  depth  of  9.5  inches  to  keep  the  bond  stress  at 
80  Ib.  sq.  in.,  and  9.0  inches  to  keep  the  shearing  stress  at  40  Ib. 
sq.  in.  As  inequalities  in  the  character  of  the  soil  may  increase 
the  unit  pressure  at  some  points,  the  depth  at  the  end  of  the 
cantilever  will  be  made  12  inches,  and  that  at  the  vertical  slab 
24  inches  as  assumed. 

The  Inner  Cantilever.  Considering  the  whole  prism  of  earth 
above  this  part  of  the  footing  to  be  supported  by  the  earth 
below  BD,  the  shear  at  B  is  10,350  +  1380  -  5750  =  5980 
pounds.  Then  the  spacing  of  £  inch  round  rods  will  be  in 
accordance  with  (2)  mo  =  V  ^  uj  =  5980  -s-  (80  X  J  X  21.5) 
=  4.0  inches.  As  the  circumference  of  the  rod  is  2.75  inches, 
m  =  4.0  -T-  2.75  =  1.45,  and  the  spacing  is  12  ^  1.45  =  8.25 
inches.  The  depth  necessary  to  keep  the  unit  shearing  stress 
at  40  pounds  per  square  inch  is  V  -r-  (35  X  12  d),  so  the  depth 
becomes  d  =  14.3  inches  which  is  less  than  that  assumed.  With 
J  inch  rods  spaced  at  8j  inches  the  per  cent,  of  reinforcement  is 
0.6013  -f-  (21.5  X  8.25)  =  0.34.  In  Plate  I  it  is  seen  that  Ni 
is  0.124,  so,  the  moment  of  the  prism  of  earth,  the  footing  and 
the  upward  pressure  under  the  footing  being  195,000  Ib.  in., 
the  required  depth  is  d2  =  195,000  -5-  (0.124  X  400  X  12)  =  328 
and  d  =  18.1  inches,  also  less  than  was  assumed.  If  d  =  18.1 


ELEMENTARY  DESIGN 


171 


inches  be  adopted,  the  spacing  of  the  reinforcement  will  be 
8.25  X  18.1  -T-  21.5  =  6.9  inches.  In  the  problem,  the  depth 
of  both  sides  of  the  footing  will  be  24  inches  at  the  vertical 
slab,  and  12  inches  at  the  extremities. 

Anchoring  the  Reinforcement.  In  this  type  of  retaining  wall 
it  is  most  important  that  the  ends  of  the  rods  be  secure  against 
slipping.  Fig.  66  shows  common  methods  of  placing  the  steel 
in  the  footing.  If  the  earth  back  of  the  wall  be  put  in  place 

before  the  outer  footing 
is  covered,  the  projection 
shown  below  the  footing 
slab  in  (6)  may  be  neces- 
sary to  prevent  sliding, 
and  it  is  often  made  to 
serve  this  purpose  and 
also  to  provide  length  of 

grip  for  the  vertical  rods.  This  is,  perhaps,  the  most  common 
design  for  footings.  The  arrangement  shown  in  (a)  may  be 
used  only  when  the  footing  is  deep  in  comparison  with  the 
height  of  the  wall.  If  the  reinforcement  be  of  large  diameters 
there  is  some  advantage  in  using  only  straight  rods,  but  sizes 
up  to  one  inch  are  readily  bent  to  any  desired  angle. 

Drainage.  As  the  pressure  of  the  earth  against  the  wall 
increases  greatly  with  the  addition  of  moisture  in  the  backing, 
it  is  well  to  provide  means  of  carrying  off  any  accumulation  of 
water  that  may  tend  to  exist  back  of  the  wall.  This  is  done 
by  means  of  drain  pipe  extending  through  the  wall  just  above 
the  level  of  the  ground,  as  shown  in  Fig.  66  (a) . 

Forms.  A  design  for  a  form  for  this  variety  of  wall  is  shown 
in  Fig.  68.  This  design  was  made  by  the  engineers  of  the 
Isthmian  Canal  Commission  for  a  reservoir  wall. 

Sequence  of  Work  in  Design  of  Cantilever  Retaining  Wall 

General  Instructions.  Size  of  sheet  is  22"  X  30"  —  border 
28"  X  30". 

Scale  of  drawing  is  1  inch  =  4  feet. 

The  cross-section  is  in  the  upper  left  hand  corner.  Under 
this  the  diagram  of  pressures  on  the  ground,  and  the  plan  show- 
ing reinforcement  in  the  footing  below. 

In  the  upper  right  hand  corner  is  the  elevation  of  ten  feet  of 


172 


REINFORCED  CONCRETE 


--H--2- 


-f-rt- 


-f-ft-f- 


2700 


X  in.  Rods  

Spaced  5%  iuches 


Shear  Diagram 


FIG/  67. 


ELEMENTARY  DESIGN 


173 


wall,  showing  spacing  and  length  of  reinforcing  bars.  Below 
this  is  a  bill  of  material  for  ten  feet  of  wall.  The  title  is  in  the 
lower  right  hand  corner. 

Data.  Gravel  foundation.  Allowable  pressure  5  tons  per 
square  foot.  Earth  filling  back  of  wall  weighs  100  pounds  per 
cubic  foot.  The  surcharge  is  li  to  1.  The  pressure  of  the 
earth  against  the  wall  is  assumed  J  X  25  X  h2. 

Concrete  weighs  150  pounds  per  cubic  foot.  C  =  400, 
St  =  12,500,  u  =  80,  and  v  =  40  pounds  per  square  inch,  n  =  15. 
The  height  of  wall  is  — .  The  computations  are  for  one  linear 
foot  of  wall. 

Design  —  The    Vertical   Wall.     Assume    0.45  h    or    0.5  h    as 


FIG.  68. 

length  of  the  base,  and  locate  the  vertical  slab  nearly  over  the 
outer  £  point.  Assume  the  thickness  of  the  base,  as  h/W,  or 
sufficient  to  extend  below  frost. 

Assume  a  thickness  of  10  inches  to  12  inches  at  the  top 
of  the  wall  and  compute  the  thickness  at  the  top  of  the  footing. 
In  high  walls  one  or  two  thicknesses  at  intermediate  heights 
may  be  computed  and  the  batter  determined  accordingly.  About 
0.005  is  a  common  value  for  p. 


174  REINFORCED  CONCRETE 

Compute  the  weight  of  the  vertical*  wall,  the  footing,  and  the 
earth  prism.  Also  find  the  horizontal  pressure  of  the  earth 
against  the  wall,  and  the  lever  arms  of  the  weights  with  the 
^  point  as  a  center.  Compute  the  moments  of  external  forces 
about  the  outer  J  point  as  a  center,  and  change  the  length  of 
the  base  if  the  overturning  moment  be  in  excess,  or  the  safe 
earth  pressure  under  the  footing  be  exceeded. 

Determine  the  spacing  of  the  vertical  reinforcement. 

Determine  the  spacing  of  the  horizontal  reinforcement  for 
temperature  stresses,  using  p  =  0.004.  About  §  of  this  should 
be  on  the  face  of  the  wall.  Test  the  wall  at  the  top  of  the  foot- 
ing for  shear  and  bond  stresses. 

The  Outer  Cantilever.  Compute  the  vertical  pressure  on 
the  ground  under  the  footing  at  each  end,  and  draw  the  pressure 
diagram. 

Find  the  shear  under  the  inner  and  outer  edges  of  the  vertical 
wall. 

Compute  the  upward  pressures  on  the  footing  on  each  side 
of  the  middle  of  the  vertical  wall. 

Take  the  center  of  moments  under  the  middle  of  the  vertical 
wall,  and  with  an  assumed  value  of  p,  find  the  depth  of  the 
footing  for  bending,  for  shear  and  for  bond  stresses.  Change 
the  assumed  depth  if  found  necessary. 

Make  the  thickness  at  the  outer  edge  of  the  cantilever  at 
least  12  inches. 

The  Inner  Cantilever.  Make  computations  similar  to  those 
for  the  outer  cantilever  and  check  assumed  depth  for  moments, 
for  shear,  and  for  bond. 

Make  the  inner  edge  12  inches  deep,  and  connect  with  a 
computed  depth  at  the  vertical  slab. 

The  Reinforcement.  Compute  the  length  of  grip  necessary 
to  secure  the  ends  of  the  vertical  rods  according  to  one  of  the 
ways  shown  in  Fig.  66,  and  provide  hooks  if  space  for  grip 
be  limited. 

Allow  a  covering  of  1  \  to  2  inches  of  concrete  over  all  reinforce- 
ment. 

Design  of  a  Counterfort  Retaining  Wall 

It  is  not  common  to  construct  retaining  walls  of  the  cantilever 
type  higher  than  25  feet,  although  there  might  be  some  economy 


ELEMENTARY  DESIGN 


175 


6  -  X  in.  Rods 


1  -  %  in.  Rod 


c  in.  Rods 


2  -  K  in.  Rods 


in.  Rods 


\\   5  -  1%  in.  Rods 


p        .  •   I  I 

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FIG.  69. 


176  REINFORCED  CONCRETE 

in  so  doing.  In  the  wall  with  counterforts  the  vertical  slab  is  a 
horizontal  beam  fixed  to  the  counterforts  and  supporting  a 
horizontal  load.  The  inner  footing  is  also  a  simple,  rather  than 
a  cantilever  beam.  The  outer  footing  is  a  cantilever  and  the 
counterfort  is  usually  so  considered. 

The  Vertical  Slab.  The  unit  horizontal  pressure  of  the  earth 
will  be  taken,  as  before,  at  25  pounds  per  square  foot.  The 
height  is  25  feet,  and  the  base  will  be  assumed  to  be  25  X  0.45 
=  11.5  feet.  The  safe  stress  on  the  steel  will  be  16,000  Ib.  sq. 
in.,  and  for  the  concrete  500  Ib.  sq.  in.  The  values  of  bond 
and  shearing  strength  will  be  80  and  40  Ib.  sq.  in.  respectively, 
and  n  is  15.  The  safe  pressure  on  the  gound  is  assumed  to  be 
5  tons  per  square  foot. 

Assuming  the  footing  to  be  2  feet  thick  and  the  counterforts 
10  feet  apart,  center  to  center,  the  bending  moment  on  the 
slab  at  the  top  of  the  footing  will  be  for  a  breadth  of  12  inches, 
M  =  A  25  X  23  X  10  X  10  X  12  =  69,000  Ib.  in.  If  p  be 
0.005,  Plate  I  gives  69,000  =  0.144  X  500  X  12  d2,  or  d  =  9 
inches.  Adding  2.5  inches  outside  the  middle  of  the  rods,  the 
thickness  is  11.5  inches.  This  thickness  must  be  tested  for 
shear  and  for  bond.  The  shear  is  25  X  23  X  5  =  2875  pounds 
if  the  thickness  of  the  counterfort  be  neglected.  Then  v  =  2875 
-r-  (12  X  9)  =  26.7  pounds  per  square  inch,  and  the  thickness 
is  sufficient.  If  the  reinforcement  be  9  X  12  X  0.005  =  0.54 
sq.  in.,  and  made  up  of  rods  of  f  inch  diameter,  the  spacing  will 
be  12  X  0.307  -r-  0.54  =  6.8  inches.  To  make  the  bond  stress 
80  Ib.  sq.  in.,  d  =  2875  X  6.8  -f-  (1.964  X  J  X  12  X  80)  =  11.8 
inches.  So  the  thickness  at  the  top  of  the  footing  is,  with  the 
covering  outside  the  rods,  14  inches  instead  of  11.5  inches  re- 
quired by  the  bending  moment.  As  the  shear  varies  directly 
with  the  depth  of  earth,  the  spacing  is  increased  accordingly. 
At  a  third  of  the  height  of  the  wall  above  the  footing  the  spacing 
is  6.8  X  1.5  =  10.2  inches.  As  the  temperature  stresses  increase 
toward  the  top  the  spacing  for  the  upper  two-thirds  of  the  height 
may  be  kept  uniform.  The  slab  may  be  of  uniform  thickness 
if  the  saving  in  expense  of  forms  will  be  more  than  the  cost  of 
the  added  concrete.  In  this  design  the  top  will  be  made  10 
inches  and  the  batter  on  the  back  will  be  uniform.  At  the 
supports  the  bending  moment  is  negative  and  numerically  equal 
to  the  positive  moment  at  the  middle  of  the  span.  The  moment 


ELEMENTARY  DESIGN  1  77 

is  negative  for  about  one-fifth  of  the  span  on  each  side  of  the 
counterfort,  and  the  outer  reinforcement  may  be  bent  inward 
to  carry  it.  The  expense  of  bending  the  rods  and  fixing  them 
in  position  is  sometimes  more  than  that  of  providing  other  rods 
of  similar  size  and  spacing  over  the  counterforts.  The  latter 
method  will  be  adopted  in  this  problem. 

As  the  unit  shear  does  not  exceed  40  pounds  per  square  inch 
no  diagonal  reinforcement  is  required. 

The  Footing.  The  length  of  the  outer  cantilever  will  be 
assumed  to  be  3  feet.  The  weight  of  the  vertical  slab  is  J  (10  + 
14)  X  23  X  150  -f-  12  =  3450  pounds.  The  prism  of  earth  with 
the  1}  to  1  slope  weighs  V  (7.33  +  7.66)  100  +  J  X  7.66  X  5.1 
X  100  =  19,210  pounds,  and  the  footing,  if  2  feet  thick,  weighs 
11.5  X  2  X  150  i=  3450  pounds.  The  center  of  gravity  of  these 
forces  is  found  to  be  6.92  feet  from  E  (Fig.  65).  The  horizontal 
pressure  per  foot  of  length  of  the  earth  backing  is  P  =  |  X  25 
X  25  X  25  =  7812  pounds  applied  8.33  feet  above  the  base. 
From  similar  triangles,  the  resultant  force  is  found  to  pierce 
the  base  0.6  feet  inside  the  middle  third  of  the  length,  or  1.32 
feet  from  the  middle,  so  the  assumption  as  to  the  length  of  base 
was  correct  if  the  allowable  pressure  on  the  ground  be  not 
exceeded.  The  pressure  on  the  earth  at  E,  Fig.  65,  is 
26110  6  X  1.32  X  26110 


L5  11.5X11.5 

and,  at  D,  710  pounds  per  square  foot.     This  is  well  within 
the  assumed  bearing  power  of  the  earth  under  the  footing. 

The  weight  of  the  wall  and  of  the  prism  of  earth  is,  for  assumed 
dimensions,  26,100  pounds.  The  horizontal  pressure  is  7812 
pounds  or  0.30  of  the  weight.  To  prevent  sliding  the  coefficient 
of  friction  should  be  considerably  more  than  0.30.  Usually  the 
footing  is  in  a  trench  dug  for  the  purpose  and  sliding  can  only 
take  place  when  the  wall  is  forced  into  this  solid  bank  of  earth. 
If  the  wall  be  built  on  top  of  ground,  the  covering  of  the  footing 
to  be  done  later,  projections  into  the  earth  under  the  footing 
are  provided.  This  projection  may  be  at  the  rear  extremity 
of  the  footing  and  may  be  used  to  lengthen  the  grip  of  the  ten- 
sion rods  in  the  counterforts,  or  it  may  be  built  where  the  ground 
is  most  solid,  or  under  the  center  of  gravity  of  the  loads.  In 
this  problem  it  is  assumed  that  no  such  projection  is  needed  to 
prevent  sliding. 


178  REINFORCED  CONCRETE 

The  Outer  Cantilever.  The  reinforcement  may  be  assumed 
upon  the  basis  of  p  =  0.005  for  the  depth  of  2  feet.  If  J  inch 
rods  be  used  the  spacing  will  be  0.6013  ^  (22  X  0.005)  =  5.5 
inches.  The  depth,  as  determined  by  the  bond,  will  be  d  =  V  -f- 
70  mo  =  (3830  +  2990).  1.5  ^  70  X  2.75  X  2.2  =  24.2  inches 
to  the  middle  of  the  steel.  For  shear,  d  =  10,230  -f-  (12  X  35) 
=  24.4  inches.  For  sufficient  bending  strength  d2  =  M  -f  NiCb. 
The  center  of  moments  will  be  taken  under  the  middle  of  the 
vertical  wall  as  in  the  previous  problem.  The  upward  pressure 
increases  from  2820  pounds  under  the  wall  to  3830  at  the  outer 
end,  and  the  lever  arm  of  the  resultant  is  found  to  be  3.6  (2820  + 
3830  X  2)  -^  3  (2820  +  3830)  =  1.89  feet.  Using  Plate  I, 
d2  =  3325  X  3.6  X  1.89  ^  (0.144  X  500)  =  315  square  inches, 
and  d  =  17.8  inches  to  the  center  of  the  steel.  Hence,  the  thick- 
ness at  the  vertical  wall  will  be  taken  as  the  greatest  of  these 
three  determinations,  or  24.4  +  2.5  =  26.9  inches.  At  the  outer 
extremity  the  thickness  may  be  reduced  to  12  inches.  It  is 
well  to  avoid  sharp  reentrant  angles  on  either  side  of  the  vertical 
slab,  and  the  forms  may  be  so  built  as  to  avoid  them  without 
much  extra  expense.  The  rods  will  not  be  stressed  to  their 
limit  in  tension,  so  the  grip  will  be  sufficient  if  they  are  not 
formed  into  books  in  front,  while  more  than  25  diameters  are 
available  in  the  rear. 

The  Inner  Floor  Slab.  In  this  type  the  inner  slab  is  a  con- 
tinuous beam  partially  fixed  at  the  counterforts  which  form 
the  supports.  In  the  design  of  the  outer  cantilever  the  slight 
downward  pressure  of  the  covering  of  earth  was  neglected,  but 
here  the  resultant  of  the  upward  and  downward  forces  must 
be  considered.  As  the  former  pressure  decreases  and  the  latter 
increases  with  the  distance  from  the  vertical  slab,  the  resultant 
is  maximum  at  the  inner  extremity  of  the  footing  and  so  a  strip 
one  foot  wide  at  that  point  will  be  considered. 

The  average  upward  pressure  on  this  strip  is  845  Ib.  linear 
in.,  the  average  depth  of  earth  over  the  back  of  the  footing  is 
26.1  feet  and  the  corresponding  weight  of  a  prism  of  a  square 
foot  of  horizontal  section  is  2610  pounds.  To  this  is  added  the 
weight  of  2  cubic  feet  of  masonry  forming  the  footing,  so  the 
uniform  load  on  the  beam  is  2610  +  300  -  845  =  2065  pounds 
per  linear  foot  of  beam  12  inches  wide.  The  length  of  the  beam 
will  be  taken  as  the  distance  between  the  middle  of  the  counter- 


ELEMENTARY  DESIGN  179 

forts  or  ten  feet.  M  =  yV  X  2065  X  100  X  12  =  247,800  =  Ni 
X  500  X  12  <f,  and  Nid2  =  41.7  square  inches.  If  p  =  0.005, 
Ni  =  0.144,  and  d  =  17.0  inches  to  the  middle  of  the  steel. 
The  shear  at  the  edge  of  the  counterfort,  assuming  the  latter 
to  have  a  width  of  16  inches,  is  8920  pounds,  and  d  =  8920  -r- 
(12  X  35)  =  21.3  inches.  For  bond  stresses  of  80  pounds  per 
square  inch  on  f  inch  round  rods  spaced  5  inch  c.  to  c.  the  depth 
is  8920  -f-  (80  X  2.4  X  2.75)  =  17.0  inches.  The  depth  at  the 
counterfort  will,  hence,  be  21.3  +  2.5  =  23.8  inches  as  required 
for  proper  shearing  stress.  This  depth  might  be  reduced  to 
20.5  inches  at  places  where  the  shear  is  sufficiently  decreased, 
but  sometimes  the  saving  of  concrete  does  not  equal  the  increased 
price  of  construction,  and  the  depth  will  be  made  uniform.  The 
spacing  of  the  rods  will  be  increased  toward  the  vertical  slab. 
At  3  feet  from  the  end  the  shear  is  5200  pounds,  and  the  spacing 
is  increased  to  5  X  9035  -5-  5200  =  8.5  inches;  at  4  feet  from 
the  end  the  shear  is  4800  pounds  and  the  spacing  is  10.7 
inches. 

The  bending  moment  is  negative  at  the  supports  and  reinforce- 
ment equal  to  that  on  the  under  side  is  inserted  at  the  top  and 
extended  ^  of  the  clear  span  on  each  side  of  the  counterfort. 

The  Counterfort.  The  vertical  slab  is  kept  in  an  upright 
position  by  the  counterforts  which  may  be  considered  as  can- 
tilever beams,  fixed  to  the  footing,  and  supporting  the  horizontal 
pressure  of  the  earth  backing.  Each  counterfort  will  then 
carry  the  load  against  10  feet  of  wall,  and  the  bending  moment 
resulting  is  5,926,000  Ib.  in.  The  depth  of  the  beam  on  a  line 
through  the  vertical  slab  at  the  top  of  the  footing  and  perpen- 
dicular to  the  back  edge  of  the  counterfort  is  90  inches  to  the 
reinforcement.  Then  from  Plate  I,  5,926,000  =  0.144  X  500  X 
b  X  8100  and  b  =  10.2  inches.  Then  A  =  0.005  X  90  X  10.2 
=  5.6  sq.  in.  This  area  will  be  supplied  by  five  lfV  inch  round 
rods,  and  the  thickness  of  the  counterfort  must  be  3  +  2.5  X  4 
X  1.2  =  15  inches.  At  15  feet  below  the  top  the  bending  mo- 
ment is  50  X  153  =  168,750  Ib.  in.,  and,  at  a  point  8  feet  higher, 
the  moment  is  50  X  73  =  17,150  Ib.  in.  The  effective  depths  of 
the  counterfort  at  these  points  are  56  'and  26  inches  respectively, 
and  Ni  -I-  r  may  be  found.  At  15  feet  below  the  top  Ni  -f-  r  = 
0.0021,  and,  by  trial  in  Plate  I,  p  is  found  to  be  0.0023,  which 
corresponds  to  a  steel  cross-section  of  16  X  56  X  0.0022  =  2.06 


180  REINFORCED  CONCRETE 

square  inches.  Two  of  the  lA  inch  rods  will  be  sufficient, 
and  these  will  be  continued  to  the  top. 

As  the  concrete  is  assumed  to  be  incapable  of  supporting 
tensile  stresses  the  counterfort  must  be  bonded  to  the  vertical 
and  footing  slabs  by  reinforcing  rods.  The  force  tending  to 
part  the  slabs  and  the  counterfort  is  the  shear  along  the  hori- 
zontal and  vertical  edges  of  the  latter.  At  the  top  of  the  footing 
the  shear  on  one  foot  in  height  is  25  X  23  X  8.75  =  5050  pounds. 
If  half-inch  round  rods  be  used  the  number  in  12  inches  of  height 
of  wall  will  be  5050  -s-  (0.196  X  16,000)  =  1.6.  These  rods  are 
inserted  in  pairs,  each  hooked  around  the  outer  horizontal 
reinforcing  bars  and  extending  back  50  diameters  into  the  coun- 
terfort. The  spacing  of  these  pairs  at  the  footing  will  be  12  -f- 
0.8  =  15  inches.  At  20  feet  below  the  top  the  shear  on  12 
inches  of  height  is  4375  pounds;  at  15  feet,  3300  pounds;  at 
10  feet,  2200  pounds,  and  at  5  feet,  1100  pounds.  The  spacing 
of  the  pairs  of  bonding  rods  between  these  limits  will  accordingly 
be  17,  22,  and  34  inches.  The  arrangement  of  the  twelve  pairs 
of  rods  is  shown  in  Fig.  69. 

In  a  similar  manner  the  necessary  bond  between  the  counter- 
fort and  the  footing  slab  is  determined.  At  D,  Fig.  65,  the 
downward  tension  on  the  counterfort  caused  by  the  shear  on 
both  sides  is  17,840  pounds  on  12  inches  of  the  length  DE. 
On  the  succeeding  spaces  12  inches  wide  the  tension  is  decreased 
2960  pounds  per  foot.  At  the  extremity  the  tension  is  taken 
by  the  diagonal  rods  which  are  carried  to  the  under  side  of 
the  slab  and  there  bent  toward  the  front.  The  tension  at  2| 
feet  from  D  is  11,900  pounds  and  two  J-inch  round  rods  are 
provided.  They  extend  upward  to  the  batter  of  the  counterfort 
and  downward  to  hook  around  the  lower  reinforcement  of  the 
slab.  At  4J  feet  from  D  the  tension  is  5980,  and  two  f  inch 
rounds  are  sufficient.  At  6|  feet  from  the  end  one  J  inch  rod 
is  inserted  and  secured  as  just  described. 

The  Reinforcement.  The  rods  parallel  to  the  batter  of  the 
counterfort  are  bent  over  to  form  hooks  at  the  upper  ends, 
and,  at  the  lower  ends,  they  may  be  bent  into  a  horizontal 
position  by  a  curve  toward  the  front.  In  some  designs  sufficient 
anchorage  is  secured  by  extending  the  footing  and  the  rods 
downward  without  bending  the  latter  into  hooks.  In  most 
walls  of  this  type  the  length  to  be  reinforced  is  greater  than 


ELEMENTARY  DESIGN  181 

that  of  the  rods,  and  the  latter  must  be  overlapped  in  order 
that  the  tension  may  be  continuous.  If  the  beam  be  continu- 
ous and  separate  rods  are  provided  for  the  negative  moment 
the  rods  may  be  joined  where  they  are  not  in  tension  and  no 
overlapping  is  required.  If  the  same  rods  be  bent  up  to  take  the 
negative  moment  the  splicing  may  best  be  arranged  near  the 
points  of  inflection. 

In  most  designs  there  is  sufficient  length  of  grip  to  keep  the 
bond  stresses  within  proper  limits  even  without  hooked  ends 
if  the  estimated  load  be  not  exceeded.  In  case  of  failure  due 
to  unforeseen  loads  the  hooks  serve  to  render  collapse  less  sud- 
den, although  they  may  be  entirely  inert  under  working  loads 
and  ordinary  conditions. 

Sequence  of  Work  in  Design  of  a  Counterfort  Retaining 

Wall 

General  Instructions.  Size  of  drawing  sheet,  22"  X  30"; 
border,  20"  X  28". 

Scale  of  drawing,  1  inch  =  4  feet. 

Drawing  covers  two  counterforts  and  included  bay.  Plan 
is  in  lower  left  hand  corner,  section  is  above  it.  Rear  elevation 
in  upper  right  hand  corner.  The  title,  data,  and  bill  of  mate- 
rials are  in  the  lower  right  hand  corner.  Show  sizes  of  all  rein- 
forcing steel  and  indicate  the  location  of  bends  and  hooks. 

Data.  Gravel  foundation  capable  of  sustaining  5  tons  per 
square  foot.  Earth  filling  back  of  wall  having  surcharge  of 
slope  of  1.5  to  1.  The  horizontal  pressure  of  earth  is  taken  as 
25  pounds  per  square  foot  at  the  top  and  increasing  directly 
as  the  depth.  The  weight  of  filling  is  100,  and  of  the  concrete 
150  pounds  per  cubic  foot.  The  allowable  stresses  are  St  = 
16,000,  C,=  500,  u  =  80,  and  v  =  40,  all  in  Ib.  per  sq.  in.  The 
value  of  ty  is  15. 

The  height  of  the  wall  is  -  -  feet,  and  the  counterforts  are 
spaced  —  center  to  center. 

Design.  (1)  For  trial  assume  the  base  as  0.45  h.  Assume 
the  thickness  of  the  footing  as  24  inches  for  heights  not  over 
25  feet. 

(2)  The  Vertical  Wall.  Compute  the  thickness  of  a  hori- 
zontal strip  of  the  vertical  slab  12  inches  high  at  the  top  of  the 
footing,  also  at  f  and  at  }  of  the  depth.  The  vertical  slab  is 


182  REINFORCED  CONCRETE 

considered  as  a  continuous  beam  supported  by  the  counterforts. 
The  positive  and  negative  bending  moments  are  each  -fa  wl2. 
The  front  of  the  wall  is  vertical  and  the  minimum  allowable 
thickness  at  the  top  is  10  inches.  Make  a  coping  extending 
4  inches  in  front  and  12  inches  deep. 

Compute  the  shearing  and  bond  stresses  at  the  supports  and 
change  the  thickness  as  computed  for  bending  if  necessary. 
Use  u  =  V/moj  and  V  =  35  bd.  If  diagonal  tension  anywhere 
exceeds  40  Ib.  sq.  in.  insert  diagonal  reinforcement.  Provide 
reinforcement  for  negative  moments  at  the  supports. 

Compute  the  spacing  of  the  horizontal  reinforcement. 

The  Footing.  Assume  the  length  of  the  outer  footing  from 
}  to  I,  the  assumed  length  of  the  base.  Compute  the  weights 
and  positions  of  centers  of  gravity  of  the  prism  of  earth  above 
the  back  footing,  of  the  vertical  wall  and  of  the  footing.  Com- 
bine these  weights  and  the  horizontal  pressure  of  the  earth 
and  find  where  the  resultant  pierces  the  base.  This  point  should 
be  within  the  middle  third,  otherwise  the  length  of  the  base 
must  be  changed.  Draw  the  pressure  diagram  and  see  that 
the  allowable  pressure  on  the  ground  be  not  exceeded. 

Test  the  foundation  for  safety  against  sliding. 

The  Outer  Cantilever.  Assume  the  reinforcement  to  be 
0.005  of  the  vertical  cross-section  and  made  up  of  small  rods. 
Neglect  the  downward  pressure  of  the  earth  and  of  the  slab 
itself  and  find,  from  the  pressure  diagram,  the  upward  bending 
moment  and  the  vertical  shear.  Compute  the  thickness  of  the 
cantilever  for  bending,  for  shear,  and  for  bond  stresses. 

The  Inner  Slab.  Find,  from  the  pressure  diagram,  the  bending 
moment  and  the  shear  on  the  inner  edge  of  the  slab,  taking  a 
strip  12  inches  wide.  Compute  the  thickness  according  to 
bending,  shear,  and  bond.  Adopt  the  largest  of  these  results 
as  the  depth.  Make  the  depth  uniform  and  vary  the  reinforce- 
ment according  to  the  stresses. 

The  Counterforts.  Consider  the  counterfort  as  a  cantilever 
beam  having  an  assumed  width  of  --  inches  and  a  depth  equal 
the  perpendicular  distance  from  the  front  of  the  vertical  wall 
to  the  batter  of  the  rear  of  the  counterfort. '  Compute  the 
reinforcement  necessary  along  the  sloping  edge  of  the  counter- 
fort, considering  the  bending  moment  to  be  50  h3,  at  the  base, 
at  §  h,  and  at  f  h.  Use  the  shear  on  both  sides  of  the  counter- 


ELEMENTARY  DESIGN  183 

fort  as  loads  producing  tension  between  it  and  the  horizontal 
and  vertical  slabs  and  determine  the  bonding  reinforcement 
necessary. 

The  Reinforcement.  Provide,  in  every  case,  a  length  of  grip 
such  that  the  bond  strength  beyond  any  section  equals  the 
tensile  stress  in  the  section;  or  ASt  =  moux.  If  St  be  16,000 
Ib.  sq.  in.,  x  is  about  50  diameters. 

Provide,  over  the  supports,  sufficient  reinforcement  to  carry 
the  negative  stresses.  Extend  this  reinforcement  £  of  the  clear 
span  on  each  side  of  the  counterforts. 

The  tension  rods  at  the  back  of  the  counterfort  are  anchored 
by  being  bent  into  the  horizontal  at  the  bottom  of  the  footing. 
The  upper  ends  are  bent  into  hooks.  The  other  rods  in  the 
counterfort  are  hooked  around  the  horizontal  reinforcing  rods 
in  the  footing  and-  vertical  slabs  and  extend  straight  inward 
or  upward. 

Bill  of  Material.  Make  a  complete  list  of  material  needed 
for  one  bay,  including  the  cement,  sand,  gravel,  and  broken 
stone;  length  and  diameters  of  reinforcing  rods;  lumber  for 
forms;  amount  of  excavation  and  labor. 

Estimate  of  Cost.     Make  an  estimate  of  the  cost  of  a  bay 

of  the  wall,  using  the  following  prices:  cement,  at per  bbl.; 

sand,  -    -  cu.  yd.;  gravel,  -    -  per  cu.  yd.;  broken  stone,  - 
cu.  yd.;  lumber,  at  -  -  per  M;  labor,  at  -  -  per  day;  superin- 
tendence, at ;  depreciation  and  interest . 


THE  ALGEBRAIC  INVESTIGATION  OF  AN  ARCH 

An  arch  is  somewhat  like  the  upper  chord  of  a  truss  in  that 
it  may  be  entirely  in  compression,  the  tension  chord  being  sup- 
plied by  the  earth  between  the  abutments.  If  the  abutments 
be  free  to  move  they  are  pressed  outward  by  the  thrust  of  the 
arch,  the  lower  chord  is  broken,  and  the  arch  may  fail  precisely 
as  a  truss  would  fail  under  similar  circumstances.  Under  this 
conception,  the  arch  is  more  like  a  truss  than  like  a  beam,  but 
most  of  the  points  of  difference  are  more  apparent  than  real. 

Types.  Stone  arches  may  be  built  of  blocks  resting  one 
upon  another  without  cementing  material  between  them,  and 
devoid  of  tensile  strength  as  a  whole.  Arches  of  concrete 
masonry  may  be  built  to  act  in  like  manner,  but  a  monolithic 


184 


REINFORCED  CONCRETE 


construction  is  more  common.  Again,  the  arch  may  be  con- 
sidered as  stable  only  by  gravity  action,  or  as  somewhat  elastic 
and  capable  of  regaining  its  original  shape  after  the  loading 

has  been  removed.  With  steel 
reinforcement  inserted  near  the 
intrados  and  extrados,  the  arch 
ring  is  capable  of  taking  tension 
or  compression  at  the  outer  edges, 
and  a  method  of  analysis  of 
stresses  which  takes  into  account 
the  elasticity  of  the  material  is 
usually  adopted.  Some  tests  on 
arches  show  that,  even  for  the 
FlG-  70-  voussoir  type,  the  elastic  theory 

gives  the  best  indication  of  the  stresses  resulting  from  the  loads. 
Analysis.  Let  A  BCD  in  Fig.  70  be  an  elementary  length 
of  an  arch  ring,  and  80  the  angle  between  the  radial  sections 
at  the  ends.  If  the  result  along 
the  arch  ring  be  applied  other- 
wise than  at  the  gravity  axis,  ef, 
the  angle  80  changes  by  8<#>,  being 
made  larger  or  smaller  according 
as  the  point  of  application  of  A 
the  pressure  is  below  or  above 
the  axis.  As  the  radius  of  cur- 
vature is  large  in  comparison 


FIG.  71. 


is    large   in 

with  depth  of  the  arch  ring,  bB  and  cC  may,  without  material 
error,  be  considered  equal.  Under  the  application  of  this  thrust 
the  deformation  at  a  point  distant  z  from  the  gravity  axis  is  z  8  <j>. 
Or  if  z  be  fb  =  c  the  corresponding  stress  is,  from  the  usual 
formula  for  deformation,  X  =  Pl/AE,  and  if  C  be  the  unit  stress, 


but  from  M  =  CI/c 

C  =  Mc/I  = 
from  which  &<t>  =  M&s/EI 

and  M  =  EI^ 

In  Fig.  71  AB  is  the  position  of  the  gravity  axis  of  the  arch 


ELEMENTARY  DESIGN  185 

ring  before  the  loads  are  applied.  If  it  be  free  to  move,  there 
is  a  tendency  for  A  to  take  the  position  A'  under  the  action  of 
loads  between  A  and  any  point  as  N.  The  distance  A  A'  is 
ANB<j>,  and,  from  similar  triangles,  By.  ANB<f>  =  x:  AN',  also, 
Bx:  ANB$  =  y:  AN,  or 

Bx  =  yB<j>  and  By  =  xB<}> 
but,  as  above,  B<f>  =  MBs/EI 

so  Bx  =  MyBs/EI 

and  *y  =  MxBs/EI. 

If  the  integration  be  extended  over  the  length  AB, 


X  =  JBAMyWl 

CA  *,    8s 

y  =    I     Mx-pj 
JB  til 


While  these  formulas  are  somewhat  less  strictly  true  for  small 
finite  lengths  of  arc,  s,  than  for  infinitesimal  lengths,  BS,  no  mate- 
rial error  results  if  the  fraction  BS/  I  be  considered  a  constant 
when  the  finite  lengths,  s,  vary  as  /  for  all  sections  between 
A  and  B.  Then,  E  being  constant, 


A 


=  s/EI2BM  (126) 

(127) 


(128) 

These  are  the  three  fundamental  equations  for  arch  analysis. 
The  three  additional  equations  may  be  written  from  the  condi- 
tions of  equilibrium: 

The  algebraic  sum  of  the  horizontal  forces  =  zero,  (129) 

The  algebraic  sum  of  the  vertical  forces  =  zero,  (130) 

The  algebraic  sum  of  the  moments  at  any  point  =  zero.  (131) 

These  six  equations  are  sufficient  for  the  solution  of  arches 
of  the  three  types  in  common  use. 

Arches  are  built  having  (a)  hinges  at  each  support,  and 
at  the  crown;  (b)  hinges  at  the  supports  only;  and  (c)  hingeless. 


186 


REINFORCED  CONCRETE 


The  Three-hinged  Arch,  (a)  In  Fig.  72  the  resultant  of  any 
load,  P,  on  a  three-hinged  arch  must  pass  through  definite 
points  at  A,  C,  and  B  on  the  arch  axis,  and,  there  being  no 


FIG.  72. 

eccentricity  of  the  point  of  application,  there  can  be  no  moments 
at  these  points.     Then,  taking  the  moments  about  B,  A,  and  C, 

Vil  -  P  (1/2  +  x)  =  0 
V2l  -  P  (1/2  -  x)  =  0 
H&  +  Px  -  Vil/2  =  0 
Vl  +  72  -  P  =  0 
Hi  -  H2  =  0. 

If  the  loads  be  inclined,  the  same  equations  result  if  hori- 
zontal and  vertical  components  be  used  instead  of  P.  With 
the  horizontal  and  vertical  components  of  the  reactions  and 


FIG.  73. 

the  points  of  application  known,  the  shear  and  the  bending 
moments  may  be  readily  determined  at  any  point. 

The  Two-hinged  Arch,  (b)  In  Fig.  73  the  two-hinged  arch 
is  free  to  rotate  at  A  and  B,  but  is  otherwise  fixed  at  these 
points.  Hence,  equations  (127)  and  (128)  become 


ELEMENTARY  DESIGN  187 


2    My  =  OandSMz  =  0 

also  Hi  -  #2  =  0 

Vil  -  P  (1/2  +  x)  =  0 
7i  +  F2  -  P  =  0 

These  equations,  modified  as  in  (a)  for  inclined  loads,  may  be 
solved  for  H\,  H2,  Vi,  and  ¥2,  after  which  the  moment,  thrust, 
and  shear  at  any  section  may  be  found. 

The  Hingeless  Arch,  (c)  In  the  hingeless  arch,  Fig.  74,  the 
points  of  application  of  the  resultant  on  the  right  or  on  the 


FIG.  74. 

left  are  unknown,  and  there  are  usually  bending  moments, 
M2  and  Mi  at  these  points.  There  are  then,  at  each  support, 
three  unknown  quantities,  or  six  in  all,  MI,  M%,  HI,  H2,  Vi, 
and  Vzt  to  be  found  and  all  of  equations  (126)-(131)  are  needed. 
Since  the  arch  is  fixed  at  A  and  B  there  is  no  change  in  the 
direction  of  the  tangent  to  the  curve  of  the  gravity  axis  of  the 
arch  ring  at  these  points,  so  (126)  becomes  zero.  Also,  accord- 
ing to  definition,  A  and  B  do  not  change  relatively  as  to  height 
or  distance  apart,  so  (127)  and  (128)  also  become  zero.  In 
the  above  formulas  M  represents  the  resisting  moment  in  the 
arch  ring,  and,  to  be  of  use  in  the  analysis  or  stresses,  must  be 
expressed  in  terms  of  the  moments  of  the  loads. 

In  Fig.  75  the  arch  ring  is  separated  at  the  crown  and  the 
forces  are  replaced  by  their  resultant,  TC)  which  is  resolved 
into  components,  Hc  and  Ve.  For  the  right  half  of  the  arch 
the  moment  at  the  crown,  Mc,  is  positive  when  Tc  is  applied 
above  the  gravity  axis,  and  Vc  is  positive  when  Tc  is  inclined 
upward  as  shown,  otherwise  Mc  and  Vc  are  negative.  If  mr 
be  the  moment  of  loads  between  the  crown  and  any  point,  xy} 


188 


REINFORCED  CONCRETE 


on  the  right  of  the  crown,  Mr,  the  resulting  moment  at  that 
point,  and  Mc,  the  bending  moment  at  the  crown  Hce 


and 


Mr  =  Mc  +  mr 
MI  =  Mc  +  ml 


Hcy 


(132) 
(133) 


The  signs  of  all  moments  producing  compression  in  the  upper 
fibers  of  the  arch  ring  are  positive,  and  each  half  of  the  arch  is 
considered  as  a  cantilever  beam.  Hence  m  is  always  negative, 
Mc  has  like  signs,  and  Vc  opposite  signs  on  the  right  and  left 
sides  of  the  crown. 

Since  in  Fig.  71  A  is  any  point  on  the  gravity  axis  of  the  arch, 
it  may  be  taken  at  the  crown  and  the  limits  of  the  summations 
in  (126),  (127),  and  (128)  will  be  over  half  the  arch.  Then 


FIG.  75. 

the  summation  on  either  side  of  the  side  of  the  crown  will  be 
numerically  equal  in  (126)  and  (127),  but  with  opposite  signs. 

Letting     M/    and    M  r    indicate  2  c  M  and  2  c  M  respectively 
in  Fig.  74,  there  may  be  written  from  (126),  (127),  and  128), 

2  MI    =  -  S  Mr  (134) 

2Mty  =  -  ^Mry  (135) 

S  MIX  =       S  MrX  (136) 


The  above  summations  are  obtained  as  follows:  the  arch 
ring  is  divided  into  parts  such  that  s/I  is  constant,  and  the 
load  and  point  of  application  of  the  same  are  computed  for 
each  division.  The  moments  of  all  forces  between  each  load 
and  the  crown  are  computed  about  the  point  of  application  of 
each  load,  and  the  results  are  added. 

After  substituting  the  values  given  in  (132)  and  (133)  the 
following  equations  result: 


ELEMENTARY  DESIGN  189 

2  nMc  +  2  #c%  +  2mz  +  2mr  =  0  (137) 

2  Mc%  +  2  #CV  +  Sw#  +  2mr?/  =  0  (138) 

2  7c2z2  +  2mzz  -  2mrz  =  0.  (139) 


The  moment  Mc  is  constant  for  a  given  loading,  but  in  the 
summation  it  occurs  once  for  each  of  the  loads  considered, 
and,  n  being  the  number  of  loads  on  each  side  of  the  crown, 
=  nMc. 

From  (139) 


From  (137)  and  (138) 

„.  _  n  (S  m/  y  +  S  mr  y)  -  (S  m/  +  S  mr)  S  y  _  n       ,       , 


+  a  mry)  -  (S  w/  +  S  wr)  Si/2 


These  equations  are  sufficient  for  the  computation  of  the 
stresses  at  any  point  in  the  arch  ring  when  the  loads  and  the 
points  of  application  are  known. 

The  distances  x  and  y  are  measured  from  a  scale  drawing  of 
the  arch,  and  the  moments  of  loads  are  readily  computed  alge- 
braically, or  they  may  be  found  graphically  as  will  be  shown 
below.  For  this  purpose  the  equilibrium  polygon  will  be  so 
drawn  as  to  be  available  for  the  algebraic  as  well  as  for  the 
graphic  method  of  solution,  and  it  need  not  be  repeated.  (See 
page  194). 

Stresses  at  any  Section.  After  the  numerical  values  of  Ve, 
Hc,  and  Mc  have  been  found,  the  resultant  pressure,  T,  at  any 
section  is  best  found  from  the  force  polygon.  The  component 
of  T  in  the  direction  of  the  radial  section  is  the  shear,  and  that 
normal  to  the  section  is  the  thrust  at  that  section.  Then  the 
bending  moment  at  any  section  is  given  by  formulas  (132)  and 
(133). 

Having  the  bending  moments  given,  the  stresses  in  a  rein- 
forced concrete  arch  ring  are  obtained  by  applying  the  prin- 
ciples on  pages  90  to  100. 

Stresses  Due  to  Temerature.  A  rise  or  fall  in  temperature 
tends  to  cause  a  change  in  the  length  of  the  span  of  an  arch. 
As  the  arch  proper  is  much  more  susceptible  to  such  changes 


190 


REINFORCED  CONCRETE 


than  is  the  earth  or  rock  upon  which  the  abutments  rest,  there 
will  be  a  relative  as  well  as  an  absolute  change  in  the  lengths 
of  span  of  the  arch  and  of  that  between  the  abutments.  In 
the  analysis  of  stresses  it  may  be  assumed  that  the  distance 

between  the  abutments 
remains  constant.  In  an 
arch  with  hinges  the 
temperature  stresses  are 
nearly  zero,  as  the  curve 
is  free  to  change  direction 
at  two  or  three  points, 
but  in  the  hingeless  type, 
the  curve  assumes  the  dotted  positions  in  Fig.  76  only  by  resist- 
ing bending  at  A  and  B,  and  creating  stresses  along  the  entire 
curve.  If  e  be  the  coefficient  of  expansion  of  concrete,  t  the 
change  in  temperature  from  the  normal,  and  I  the  span,  the 
change  in  length  of  the  span  will  be,  if  not  resisted,  tel.  This 
is  2<x  in  (127).  Since  there  can  be  no  change  in  the  direction 
of  the  tangents  at  A  and  B,  (126)  is  zero.  Hence, 
A 


FIG.  76. 


and 


Also,  since  no  external  loads  are  considered  and  the  arch  is 
symmetrical,  m  and  V  are  zero,  and  on  either  side  of  the  crown 

M  =  Mc  4-  Hcy  (143) 

combining  these  three  equations, 

2  s/EI  (Mc^y  +  #c%°)  =  tel 
and  nMc  +  Hc^y  =  0, 

ntel  IE 


from  which 

and 
also 


Hc  = 


2[n%2 - 


Mc  =  - 

M  =  Mf  4 


Hcy. 


(144) 

(145) 
(146) 


The  thrust  acts  along  the  line  kk,  Fig.  76,  which  is  %  -f-  n 
below  the  crown,  and  forms  the  closing  line  of  the  equilibrium 
polygon.  The  shear  and  normal  pressure  on  any  section  are 


ELEMENTARY  DESIGN  191 

found  by  the  resolution  of  H,  as  was  done  with  R  in  the  dis- 
cussion of  external  loads. 

The  positive  sign  in  (144)  is  used  when  t  indicates  a  rise  in 
temperature  and  the  negative  sign  when  the  change  is  downward. 

The  dimensions  of  the  arch  are  usually  in  feet,  and  so  E  is 
in  pounds  per  square  foot,  or  about  2,000,000  X  144. 

The  bending  moment  at  any  point  is  Mc  +  Hcy,  and  if  the 
above  value  of  Mc  be  known,  the  bending  moment  at  any  point 
is  given  by  the  product  of  H  and  the  distance  of  the  point  from 
kk.  Hence  it  is  seen  that,  Hc  being  positive,  all  moments 
above  the  line  kk  are  negative,  and  all  moments  below  kk  are 
positive.  For  a  fall  in  temperature,  Hc  is  negative  and  the 
signs  of  the  moments  are  reversed. 

Effect  of  Arch  Ring  Shortening  due  to  Thrust.  The  com- 
pression caused  by  the  thrust  along  the  arch  ring  would,  if  not 
resisted,  make  the  arc  shorter,  and  it  has  an  effect  similar  to 
that  due  to  a  fall  in  temperature.  The  change  in  length  of 
span  I,  due  to  an  average  stress  C,  in  the  concrete  is  Cl/E,  which 
corresponds  to  tel,  the  change  in  length  due  to  temperature 
variation,  or, 


At  the  crown        Mc  = ^  (148) 

and  in  general       M  =  Mc  +  Hcy.  (149) 

With  the  thrust  and  moment  given,  the  eccentricity  may  be 
found  and  the  stress  in  the  concrete  determined  by  the  prin- 
ciples, pages  90—100. 

The  combination  of  stresses  will  be  given  in  the  design  of 
an  arch,  page  206. 

THE  GRAPHIC  ANALYSIS  OF  A  HINGELESS  ARCH 

Principles.  In  Fig.  77,  ai,  a2  .  .  .  an,  is  the  gravity  axis  of 
the  arch  ring,  and  Ci,  Cz  .  .  .  cn  is  the  equilibrium  polygon  of 
n  loads  assumed  to  be  drawn  in  proper  position.  Then  the 
bending  moment  at  any  section,  as  as,  is  the  product  of  the 
pole  distance  of  the  force  polygon  from  which  ci,  Cz  .  .  .  cn  is 


192 


REINFORCED  CONCRETE 


constructed,  and  of  the  intercept  asc^;  or  M  =  H-a3c3.  Since 
H  is  constant  for  any  given  force  polygon,  the  fundamental 
formulas,  (134),  (135),  and  (136)  become 


=  0, 


=  0,  and 


=  0 


the  origin  of  coordinates  being  at  the  middle  of  the  span,  and 
xy  coordinates  of  any  point  on  the  line  a  i,  «3,  an. 

If  the  force  polygon  be  drawn  with  the  true  pole  distance, 
and  the  true  equilibrium  polygon  be  in  position,  the  stresses  in 


FIG.  77. 


the  arch  ring  may  be  found  for  any  point.  The  problem  then 
is  to  determine  the  true  pole  distance,  and  at  least  one  point 
on  the  arch  ring  through  which  the  true  equilibrium  polygon 
must  pass. 

For  some  system  of  symmetrical  loading  the  line  of  pressure 
represented  by  the  polygon  ci,  c2,  cn  will  fall,  everywhere,  on 
the  gravity  axis  ai,  a2,  an-  The  pressure  at  the  ends  of  the 
span,  ai,  an,  cause  reactions  at  these  points  which  may  be  re- 
solved into  vertical  and  horizontal  forces.  The  closing  line  of 
the  polygon  will  then  be  ai,  ki,  kn,  an,  and  the  position  of  ki,  kn 
is  to  be  determined. 

In  Fig.  77  let  ok  and  ck  represent  the  intercepts  between  the 
line  kikn  and  the  curve  a^,  and  polygon  CjCn  respectively, 


then 

from  above  formulas, 


ac  =  ck  —  ak 


—  ^ak  =  0 

—  'Zak-x  =  0 

—  ^ak-y  =  0 


(150) 

(151) 
(.152) 


The  position  of  the  curve  a\,  a2  .  .  .  an  is  already  determined 
and  (151)  and  (150)  are  satisfied  if  2ak  and  ^ak'x  are  each 
zero.  The  last  formula  (152)  is  satisfied  only  when  Ci  .  .  .  cn 


ELEMENTARY  DESIGN 


193 


is  drawn  from  a  force  polygon  having  the  true  pole  distance, 
as  will  be  shown  later. 

Locating  k'k'.  In  Fig.  78  ^c'k'  is  zero  and  equation  (150) 
is  satisfied  when  k'k'  is  so  drawn  that  c'ik'p'  +  c'nk'p'  =  p'bp', 
or,  otherwise,  when  the  area  of  the  trapezoid  c'\k'k'c'n  equals 
the  area  of  the  segment  c\bcn.  One  of  an  infinite  number  of 
positions  of  k'k'  then  satisfies  the  first  condition  when  o'k'  is 
the  mean  ordinate  of  c'\bc'n  and  o'c'\  =  o'c'n. 

The  second  equation  (151)  is  satisfied  when  ^ak-x  =  2ck-x 
=  0,  or,  in  other  words,  when  the  center  of  gravity  of  the  trape- 
zoid c'ik'c'n  is  on  the  vertical  line  through  the  center  of  gravity 

6 


FIG.  78. 

of  the  equilibrium  polygon  c'\bc'n.  This  may  be  readily  shown. 
Let  c'ik'p',  bp'p',  and  p'k'c'n  be  represented  by  p\,  p2  and  ps,  and 
let  the  centers  of  these  areas,  or  forces,  be  at  x\,  x2,  xs  from  the 
origin  of  coordinates.  Also,  let  Tt  and  Tp  be  the  areas  of  the 
trapezoid  and  the  equilibrium  polygon  respectively,  with  centers 
of  gravity  xt  and  xp  from  the  origin.  Then 


or 

and 

since 


Ttxt  =  Tpxp  +  (pixi  —  p2x2 

=  Tpxp  +  Zc'k'.x 
Ttxt  =  Tpxp 

Xt  =  xp 
c'k'.x  =  0  and  Tt  =  Tp. 


When  the  polygon  c'ibc'n  represents  the  gravity  axis  of  a 
symmetrical  arch,  c'  and  k'  become  a  and  k,  then  aikkan  becomes 
a  rectangle,  and  kk  is  parallel  to  aian,  Fig.  77,  and  at  a  distance 
above  it  equal  to  the  mean  ordinate  of  the  segment  a\azan  or 
to  the  area  aia^an  -f-  a\an. 


194 


REINFORCED  CONCRETE 


When  the  gravity  axis  of  the  arch  is  in  the  form  of  a  parabola 
the  distance  aik  is  f  of  the  rise.  If  the  curve  of  the  gravity 
axis  be  a  segment  of  a  circle  the  distance  a\k  =  ank  is  given  in 
the  following  table: 


R  +  S 

ai* 

R  +  S 

aik 

R  +  S 

ai* 

R  +  S 

aik 

R  +  S 

aik 

R  +  S 

aik 

.05 

.663 

.10 

.672 

.15 

.680 

.20 

.688 

.25 

.698 

.30 

.714 

.06 

.669 

.11 

.673 

.16 

.682 

.21 

.689 

.26 

.701 

.40 

.745 

.07 

.669 

.12 

.675 

.17 

.684 

.22 

.691 

.27 

.704 

.50 

.785 

.08 

.670 

.13 

.677 

.18 

.686 

.23 

.693 

.28 

.707 

— 

— 

.09 

.672 

.14 

.679 

.19 

.687 

.24 

.695 

.29 

.711 

— 

— 

When  the  gravity  axis  is  an  irregular  curve,  as  is  usually  the 
case,  the  height  of  kk  above  the  line  aian  is  determined  by 
dividing  the  span  into  an  even  number  of  spaces,  equidistant 
apart,  having  the  crown  in  the  middle  of  one  of  the  spaces,  and 
finding  the  mean  of  the  ordinates  at  each  division  point.  This 
is  the  usual  method  and  it  is  essential  that  the  sum  of  the  ordi- 
nates multiplied  by  the  length  of  a  space  does  not  differ  mate- 
rially from  the  area  of  the  segment  covered. 

To  MAKE  5  ck-y  =  2  ak-y 

The  Trial  Equilibrium  Polygon.  In  general,  the  pole  dis- 
tance is  unknown,  and  after  assuming  a  trial  pole,  a  trial  equilib- 
rium polygon  is  drawn.  From  this  trial  polygon  and  the  three 
formulas  (150),  (151),  and  (152),  the  true  pole  and  the  true 
equilibrium  polygon  are  found. 

In  Fig.  79  (6),  the  loads  are  laid  off  to  scale,  and  a  trial  pole 
is  chosen  opposite  the  point  on  the  load  line  between  the  loads 
next  to  and  on  either  side  of  the  crown.  The  trial  equilibrium 
polygon  is  then  drawn  as  shown  in  (c).  (In  general  the  trial 
polygon  may  be  drawn  with  any  assumed  pole,  but  for  use  in 
the  algebraic  solution,  pages  187-189,  the  pole  is  here  taken 
as  indicated.  If,  then,  the  line  through  the  two  points  men- 
tioned be  extended,  the  bending  moments  of  any  loads  about 
the  crown  is  given  by  the  product  of  the  intercept,  between  that 
line  and  the  equilibrium  polygon,  and  the  pole  distance.) 

Having  the  equilibrium  polygon  completed,  the  span,  Fig. 
79  (c),  is  divided  into  an  even  number  of  equal  spaces,  often 
the  same  as  that  of  the  loads,  one  of  the  spaces  being  bisected 


ELEMENTARY  DESIGN 


195 


by  the  middle  of  the  span.  The  mean  of  the  ordinates  is  laid 
off  at  the  middle  and  the  point  o'  is  thus  located.  If  any  line 
k'k'  be  drawn  through  o',  and  verticals  through  c'  and  c'n, 
the  areas  c'k'k'c'n  and  c'lc'^c'n  are  equal,  or  2c'fc'  =  0.  If; 
in  addition,  the  center  of  gravity  of  the  trapezoid  be  on  that 


FIG.  79. 


through  that  of  the  segment,  ^c'k'-x  is  zero.  The  problem,  then, 
is  to  so  proportion  c'k'  and  c'nkf  that  the  center  of  gravity  of  the 
trapezoid  falls  at  the  proper  distance  from  c'.  In  any  trapezoid, 
as  that  in  Fig.  80,  the  center  of  gravity  is  located  by  the  equations 


x  =  h 


2c 


3  (6  +  c) 


=  h 


2b 


3  (6  +  c) 


and   y  = 


ab  +  b2    ,    a 
H 


From  these  equations  it  may  be 
shown  that,  if  e  =  1/2  —  x,  c  = 
(b  +  c)  (I  +  6  e)  ^  2  I,  and  6  = 
(b  +  c)  (I  -  6  e)  +  2  I.  In  the 
problem  now  being  considered  e 
is  the  distance  from  the  center 
of  gravity  of  the  segment  en- 
closed by  trial  equilibrium 
polygon  to  the  middle  of  the 
span,  b  is  c'k'  if  e  be  on  the 


FIG.  80. 


right  of  the  middle  of  the  span,  otherwise  b  is  c'nfc'  and  c  is  c'A;'. 


196  REINFORCED  CONCRETE 

That  is,  the  larger  of  c'k'  and  c'nk'  is  on  the  same  side  of  the 
middle  of  the  span  as  is  e.     Then 

nt       I  -  6e     ,  ,,,       I  +  6e     , 

ck    =  — j —  oo  and  c^k    =  — -7 —  00 
I  I 

and  k'k'  may  be  correctly  located. 

The  distance  e  is  easily  computed  from  the  ordinates  at  the 
division  points.  The  differences  in  the  lengths  of  ordinates  at 
like  distances  from  the  middle  of  the  span  are  multiplied  by 
their  distances  from  the  middle,  the  products  are  then  summed 
and  divided  by  the  sum  of  the  ordinates.  Then,  in  Fig.  79  (6), 

e  =  [J  (05  -  o4)  +  f  (06  -  0s)  +  £  (0?  -  02)  +  I  (08  -  01)]  -*•  20 

The  ordinates,  as  o6  are  from  a6  to  c'c'n. 

This  result  gives  e  in  terms  of  spaces.  The  verticals  at  c'  and 
c'n  may  now  be  laid  off  and  k'\k'  may  be  drawn.  It  should 
be  noticed  that  this  construction  refers  to  c'c'6c'n  as  a  geomet- 
rical figure  and  not  to  the  line  of  action  of  the  loads  themselves. 
The  latter  could  be  found  by  producing  c'c'i  and  c'Bc'n.i.  to 
their  intersection. 

If  the  inflection  points,  p',  where  the  line  k'k'  crosses  the 
trial  equilibrium  polygon,  be  projected  up  to  kk  in  (a),  two 
points  through  which  the  true  equilibrium  polygon  must  pass 
are  located. 

The  True  Equilibrium  Polygon.  If  in  Fig.  79  (6)  a  line  be 
drawn  through  P',  parallel  to  k'k',  the  force  line  will  be  divided 
into  two  parts  representing  the  reactions  of  the  loads  at  the 
supports.  If  a  horizontal  line  be  drawn  through  this  point  of 
division,  and  if  P'  be  projected  vertically  upon  it,  a  new  pole 
P"  will  be  found  and  a  new  equilibrium  polygon  may  be  drawn 
having  k'k'  horizontal. 

If  this  polygon  be  drawn  in  (79)  (a)  through  pp,  *%ck  will  be 
equal  to  2M  and  ^ck-x  will  equal  2M-x,  but,  in  general,  ^ck-y 
will  not  equal  ^ak-y.  Since  y',  in  (c),  varies  as  the  pole  dis- 
tance, the  true  pole  distance,  H,  may  be  found  when  ^ak-y 
and  ^ck-y  have  been  scaled  and 

True  pole  distance:  Trial  pole  distance:  :  2ak-y:  ^c'k'-y'. 
With  this  new  pole  distance  the  true  equilibrium  polygon  is 
drawn  through  pp  in  (a),  and  the  three  conditions:   2M  =  0; 
=  0,  and  ^M-y  =  0  are  satisfied. 


ELEMENTARY  DESIGN  197 

From  this  point  the  determination  of  the  stresses  are  identical 
with  the  same  work  in  the  algebraic  solution  of  the  problem  on 
pages  184-186. 

Temperature  Stresses.  Since  the  stresses  depend  upon  Mc 
and  Hc  the  method  given  under  the  algebraic  solution  is  used. 
It  must  be  remembered  that  the  values  of  y  in  the  two  methods 
differ  by  the  rise  of  the  arch,  being  measured  downward  from 
the  crown  in  one  case  and  upward  from  the  middle  of  the  span 
in  the  other.  With  this  exception  the  same  formulas  apply 
in  both  methods. 

Stresses  Due  to  Arch  Ring  Shortening.  The  same  remarks 
as  made  regarding  temperature  stresses  apply  to  the  determina- 
tion of  stresses  of  this  character. 

Design  of  a  Reinforced  Concrete  Elastic  Arch  Ring 

Loads  and  Dimensions.  It  is  required  to  design,  and  compute 
the  stresses  in,  a  reinforced  concrete  arch  ring  of  50  feet  clear 
span  and  of  12  feet  rise,  and  having  no  hinges  either  at  the 
crown  or  at  the  springing  line.  The  load  will  be  assumed  as 
150  pounds  per  square  foot  in  addition  to  the  dead  load.  This 
loading  is  suitable  for  rather  heavy  traffic,  such  as  would  exist 
on  roads  near  large  cities.  In  case  more  definite  knowledge 
concerning  the  traffic  to  be  carried  is  available,  more  exact 
loading  may  be  used  in  the  computations.  As  a  live  load  is 
apt  to  be  concentrated  it  is  usual  to  assume  heavier  unit  loading 
for  short  than  for  long  spans.  The  loads  which  are  carried  to 
the  arch  ring  by  a  spandrel  filling  of  earth,  are  well  distributed, 
and  it  is  quite  common  to  assume  a  supposed  equivalent  uniform 
load  even  when  some  heavy  concentrated  loads,  such  as  a  road 
roller,  have  to  be  carried.  If  the  earth  filling  be  shallow,  a 
definite  rolling  live  load  should  be  imposed.  Road  rollers  and 
traction  engines  may  weigh  15  tons  and  an  electric  street  rail- 
way car  may  have  as  much  as  6  tons  on  an  axle.1  The  level 
spandrel  filling  is  taken  as  3  feet  deep  at  the  crown. 

The  intrados  is  the  arc  of  a  circle  and  the  depth,  in  inches, 
of  the  arch  ring  at  the  crown  is  computed  from  the  formula2 

t  =  Vs  +  0.1  s  +  0.005  wi  +  0.0025  w* 

1  For  a  discussion  of  moments  due  to  street  railway  cars  see  article  by  the 
author  in  Street  Railway  Journal,  April   11,  1903,  page  252. 

2  Engineering  Record,  Nov.  4,  1905,  p.  562. 


198  REINFORCED  CONCRETE 

in  which  s  is  the  span  in  feet,  we  and  wt  are  the  live  load  and 
dead  load  at  the  crown  in  pounds  per  square  foot  respectively. 
This  is  known  as  Weld's  formula  and,  while  it  does  not  take  into 
account  the  rise  of  the  arch,  is  quite  satisfactory  for  the  purpose 
of  preliminary  design.  Substituting  the  dimensions  in  this 
problem,  the  crown  thickness  is  found  to  be  13.5  inches.  The 
thickness  at  the  springing  is  from  two  to  three  times  that  at 
the  crown,  or  in  this  problem  26  inches.  The  thickness  at  the 
crown  is  taken,  for  trial,  as  13.0  inches. 

The  Curves  for  the  Intrados  and  the  Extrados.  When  the 
arch  is  comparatively  flat  the  segment  of  a  circle  is  often  used 
for  both  intrados  and  extrados  with  the  result  that  the  line  of 
pressure  will  fall  near  the  curve  midway  between  them.  A 
rather  more  graceful  appearance  is  given  if  the  radii  be  diminished 
near  the  abutments  and  the  curve  made  to  approach  the  form 
of  an  ellipse.  The  preliminary  extrados  is  a  smooth  curve 
between  points  fixed  by  the  thickness  of  the  arch  ring  at  the 
crown  at  the  haunch  and  at  the  skewback.  The  curve  midway 
between  the  intrados  and  extrados  is  the  linear  arch  which,  in 
the  computations,  is  the  presentative  of  the  masonry  ring.  This 
curve  is  on  the  gravity  axis  of  the  arch. 

Making  ds/I  Constant.  The  next  step  is  to  divide  the  arch 
ring  so  that  dsi+I  is  constant  from  the  crown  to  the  springing 
line.  The  moment  of  inertia  here  used  is  that  of  the  concrete 
cross-section  combined  with  that  of  the  steel,  or  the  transformed 
section  as  described  in  Chapter  IV,  page  90.  The  percentage  of 
steel  to  be  used  varies  with  the  ratio  of  live  to  dead  load,  being 
more  for  short  than  for  long  spans.  In  this  problem  the  steel  will 
be  assumed  as  one  per  cent,  of  the  cross-section  at  the  crown, 
one  half  at  2  inches  from  the  intrados,  and  half  at  the  same  dis- 
tance from  the  extrados.  Throughout  the  greater  part  of  the 
arch  ring  the  steel  will  be  nearly  1/10  of  the  thickness  of  the  con- 
crete from  the  edge,  and  the  diagrams  in  Chap.  IV  may  be  used. 

In  Fig.  82  the  length  CA  is  that  of  half  the  gravity  axis  of 
the  arch  while  the  curve  is  plotted  by  ordinates  equal  to  the 
transformed  moments  of  inertia  at  six  places  along  the  arch 

ring.     The  value  of  It  at  the  crown  is  j^  X  12  X  133  +  0.01  X 

C13  —  4)2 
15  X  156  X s-    -  =  2760   in.4  =  0.129    ft.4.      The    other 


ELEMENTARY  DESIGN 


199 


200 


REINFORCED   CONCRETE 


ordinates  are  found  in  like  manner.  The  length  CA  is  to  be 
so  divided  into  the  various  parts,  ds,  eight  in  this  case,  that 
each  ds  divided  by  the  ordinate  to  its  middle  point  will  be  con- 
stant. This  is  done  by  trial  by  finding  7a,  the  average  of  several 
ordinates,  and  the  average  ds,  or  CA  -f-  8.  The  length  of  the 
gravity  axis  of  half  the  arch  ring  is  found  by  scale  to  be  29.3  feet, 
so  the  average  ds  is  29.3  -r-  8  =  3.66  feet.  The  average  of  the 


FIG.  82. 

ordinates  is  found  to  be  0.488,  so  the  tangent  of  the  angle  a  is 
ds  -r-  2  Ia  or  3.66  -f-  0.976  =  3.8.  This  is  only  approximate 
since  it  is  based  upon  ordinates  spaced  equidistant  apart  while, 
in  the  final  division,  most  of  the  ordinates  will  be  taken  where 
the  arch  ring  is  thin.  Assuming  tan  a  somewhat  over  4,  and 
beginning  at  C,  the  triangles  in  Fig.  82  are  drawn  as  shown. 
The  final  diagonal  should  pass  through  A,  which  result  is  accom- 
plished by  a  few  changes  in  the  assumption  of  tan  a.  The 
intercepts  on  CA  are  scaled  and  laid  off  on  the  gravity  axis  as 
values  of  ds.  The  ordinates  in  Fig.  82  are  plotted  to  a  larger 
scale  than  is  AC  and  tan  a  is  determined  by  corresponding 
scales.  The  weight  of  the  parts  of  the  arch  ring  of  length  ds 
is  combined,  with  that  of  the  spandrel  filling  and  live  load  over 
the  horizontal  projection  of  the  same  lengths,  into  a  single  load 
on  each  ds.  At  points  where  this  load  passes  through  the  gravity 
axis  of  the  arch  ring  the  thickness  of  the  ring  is  scaled  and  the 
transformed  momen1"  ^f  inertia  is  computed.  Then  ds  -r-  It 


ELEMENTARY  DESIGN 


201 


should  be  constant.  By  scale  the  values  of  ds  are  found  to  be 
1.44;  1.76;  2.05;  2.56;  3.16;  4.16;  5.65;  and  8.52  feet  respect- 
ively. 

The  Reduced  Load  Contour.  For  convenience  in  finding  the 
points  of  application  of  the  loads,  the  depths  of  equivalent 
spandrel  and  live  loads  of  the  same  unit  weight  as  the  arch  ring 
are  found  and  plotted  above  the  arch  ring  in  Fig.  79.  The 
center  of  gravity  of  a  trapezoid  is  found  graphi- 
cally as  shown  in  Fig.  83.  The  shorter  segment 
of  each  diagonal,  as  Ao  and  Do,  are  laid  off 
on  the  longer  segments  at  c  and  b,  and  the  cen- 
ter of  gravity  of  the  triangle  obc  is  on  that  of 
the  trapezoid.  The  spandrel  filling  is  assumed 
to  weigh  100  pounds  per  cubic  foot  or  f  that 
of  the  arch  ring  while  the  live  load  is  that  of  a 
layer  of  masonry  12.9  inches  deep.  The  con- 
tour for  the  reduced  spandrel  filling  is,  at  all 
points,  a  distance  above  the  extrados  equal  to  y  of  the  depth 
of  the  actual  spandrel  filling.  The  live  load  is  assumed  to  cover 
half  the  span  and  the  contour  for  this  load  is  everywhere  12.9 
inches  above  that  for  the  spandrel  filling. 

The  Load  Diagram.  The  load  on  any  length  ds  is  found  by 
multiplying  the  area  of  the  corresponding  trapezoid,  in  square 
feet,  by  140,  the  lower  limit  of  the  trapezoid  being  the  intrados  of 
the  arch  ring.  In  this  manner  the  loads  in  pounds  are  found  to  be 


FIG.  83. 


No. 

Load 

Arm 

No. 

Load 

Arm 

No. 

Load 

Arm 

No. 

Load 

Arm 

1 

8830 

22.70 

5 

1275 

6.51 

9 

*850 

0.72 

13 

2350 

9.37 

2 

4580 

17.11 

6 

1000 

4.20 

10 

1060 

2.32 

14 

3220 

12.78 

3 

2660 

12.79 

7 

800 

2.32 

11 

1310 

4.20 

15 

5300 

17.09 

4 

1875 

9.38 

8 

630 

0.72 

12 

1650 

6.51 

16 

9760 

22.68 

The  arm  in  this  table  is  the  distance  from  each  load  to  the 
middle  of  the  span.  Before  drawing  the  force  or  equilibrium 
polygons  it  should  be  ascertained  whether  or  not  ds  -r-  /  is  con- 
stant. To  do  this  the  thickness  of  the  arch  ring  is  scaled  at 
each  load  and  the  transformed  moment  of  inertia  found.  These 
moments,  together  with  the  values  of  ds,  are  given  in  the  follow- 
ing table: 


202 


REINFORCED  CONCRETE 


Load 

ds 

'l 

Ratio 

Load 

ds 

h 

Ratio 

1 

8.52 

0.911 

9.35 

5 

2.57 

0.276 

9.32 

2 

5.65 

0.605 

9.34 

6 

2.05 

0.220 

9.32 

3 

4.16 

0.444 

9.37 

7 

1.74 

0.187 

9.33 

4 

3.17 

0.340 

9.32 

8 

1.44 

0.154 

9.35 

As  both  ds  and  It  are  found  by  scale,  and  therefore  subject  to 
some  error,  the  ratio  is  fairly  exact. 

The  Trial  Equilibrium  Polygon.  With  the  magnitude  and 
position  of  each  load  known,  the  force  and  trial  equilibrium 
polygons  may  be  constructed.  The  pole  distance  is  assumed 
at  20,000  pounds  and  the  pole  is  opposite  the  point  on  the  load 
line  between  the  8th  and  9th  loads.  The  loads  are  assumed  to 
act  vertically  so  the  load  line  is  straight  as  shown  in  (6).  The 
equilibrium  polygon  is  tangent  to  the  line  of  resistance  and  hence 
for  a  continuous  load,  coincides  with  that  curve  only  at  the 
points  of  tangency.  For  this  reason  ordinates  through  the  trial 
equilibrium  polygon  at  points  other  than  those  just  established 
under  the  loads  should  be  used.  In  this  problem  the  span  of 
the  trial  equilibrium  polygon,  which  is  that  of  the  gravity  axis 
of  the  arch  ring,  is  divided  into  16  equal  parts  so  that  one  part 
is  bisected  by  the  point  at  the  crown  of  the  arch.  As  these 
division  points  do  not  fall  upon  those  under  the  assumed  loads, 
the  area  of  the  polygon  is  given  with  sufficient  accuracy  by  the 
product  of  the  sum  of  the  ordinates  by  the  length  of  one  of 
the  equal  spaces.  The  ordinates  are  scaled  and  recorded  in  the 
table  in  the  second  column.  The  mean  of  these  ordinates,  8.37 
feet,  is  laid  off  on  a  vertical  through  the  middle  of  the  span  to 
locate  one  point,  o',  through  which  k'k'  must  pass  to  satisfy  the 
condition  that  S  c'k'  =  zero.  The  values  in  column  3,  below,  are 
the  differences  between  ordinates  at  like  distances  from  o',  as 
0i6  —  0i ;  015  —  02  and  so  forth.  The  sum  of  the  product 
of  each  difference  by  the  respective  distances  to  the  middle 
divided  by  the  sum  of  the  ordinates  is  the  distance  of  the  center 
of  gravity  of  the  polygon  from  the  middle.  Thus,  the  eccentricity, 
e  =  51.57  -f-  16  (13  X  J  +  35  X  I  +  50  X  f  +  55  X  J  +  50  X 
1  +  45  X  V  +  28  X  V  +  3  X  ¥)  -s-  13393  =  0.252  feet  to  the 
right  of  the  middle.  Then  c'k'  =  8.37  (51.74  -  6  X  0.252)  -^ 


ELEMENTARY  DESIGN 

51.74  =  8.12    feet   on   the   left,    and   c'fc'  = 
0.252)  ^  51.74  =  8.61  feet  on  the  right. 


203 
L37  (51.74  +  6  X 


No. 

O 

OT-OI 

m 

y 

k 

m-y 

k-y 

1 

1.85 

-6.25 

1.72 

-6.60 

-  10.76 

-  11.35 

2 

4.70 

-3.50 

4.62 

-  3.64 

-  16.17 

-  16.80 

3 

6.92 

-  1.30 

6.84 

-  1.40 

-    8.90 

-    9.58 

4 

8.53 

+  0.28 

8.60 

+  0.36 

+    2.41 

+    3.10 

5 

9.80 

1.53 

9.92 

1.68 

15.18 

16.67 

6 

10.70 

2.37 

10.88 

2.66 

25.80 

28.95 

7 

11.35 

3.00 

11.52 

3.32 

34.56 

38.30 

8 

11.72 

3.35 

11.82 

3.62 

39.60 

42.80 

9 

11.85 

13 

3.40 

11.82 

3.62 

40.10 

42.80 

10 

11.70 

35 

3.18 

11.52 

3.32 

23.60 

38.30 

11 

11.20 

50 

2.72 

10.88 

2.66 

29.60 

28.95 

12 

11.35 

55 

1.85 

9.92 

1.68 

18.33 

16.67 

13 

9.03 

50 

0.55 

8.60 

0.36 

4.73 

3.10 

14 

7.37 

45 

-  1.14 

-6.84 

-  1.40 

-    7.80 

-    9.58 

15 

4.98 

28 

-3.40 

4.62 

-3.64 

-  15.70 

-  16.80 

16 

1.88 

3 

-6.64 

1.72 

-6.60 

-  11.43 

-  11.35 

133.93 

0.00 

131.84 

0.00 

176.15 

184.18 

With  the  line  k'k'  drawn,  the  intercepts  c'k'  are  scaled  and 
tabulated  in  column  m.  The  values  of  y  are  the  ordinates  from 
the  chord  to  the  curve  of  the  gravity  axis  in  (a).  The  mean 
value  of  y,  8.24  feet,  is  the  distance  of  kk  above  the  chord  of  the 
gravity  axis  and  kk  is  drawn  parallel  to  this  chord.  The  values 
in  column  k  are  the  intercepts  ck  in  (a).  The  ordinates,  m,  vary 
inversely  as  the  pole  distance  and  therefore  the  sums  of  the  last 
two  columns  are  made  alike  by  a  suitable  change  in  the  assumed 
pole  distance  or 

True  pole  distance:  20,000  =  176.15  :  184.18 

True  pole  distance  =  19150  pounds. 

Through  the  trial  pole  a  line  is  drawn  parallel  to  the  chord  of  the 
trial  equilibrium  polygon  c'c'  to  intersect  the  load  line  at  g.  The 
true  pole  distance  is  then  laid  off  on  a  horizontal  line  through 
g  to  locate  the  true  pole,  or  in  general,  on  a  line  parallel  to  the 
chord  of  the  gravity  axis. 

The  True  Equilibrium  Polygon.  If  the  inflection  points  pf 
be  projected  upward  to  the  line  kk  in  (a),  two  points  p  will  be 
fixed,  through  which  the  true  equilibrium  polygon  must  pass. 
Beginning  at  either  point  p,  the  true  equilibrium  polygon  is 


204  REINFORCED  CONCRETE 

drawn  parallel  to  the  corresponding  rays  through  the  true  polo 
and  the  proper  points  on  the  load  line.  This  polygon  should 
pass  through  the  other  point  p,  which  fact  serves  as  a  check  on 
the  accuracy  of  the  work.  The  true  equilibrium  polygon  is 
usually  drawn  whether  the  solution  is  by  the  algebraic  or  by 
the  graphic  method.  At  the  points  where  the  equilibrium 
polygon  crosses  the  gravity  axis  of  the  arch  there  is  no  bending 
due  to  loads,  while  the  bending  moments  are  everywhere  given 
by  the  product  of  the  component  of  the  thrust  perpendicular 
to  a  section  and  the  intercept  between  the  polygon  and  the  axis 
on  the  same  section.  The  same  result  is  obtained  from  the 
product  of  the  true  pole  distance  and  the  vertical  intercept  at 
the  same  section. 

Stresses  Due  to  Loads.  In  this  problem  the  eccentricities 
of  the  polygon  are  scaled  on  radial  lines  at  the  16  points  of  even 
division  of  the  span  and  tabulated  on  page  203.  The  thrusts 
are  scaled  from  the  force  polygon,  the  difference  between  those 
on  the  right  and  on  the  left  being  due  to  the  difference  in  loads. 
The  bending  moment  at  any  point,  as  11,  is  then  19,500  X  0.2  = 
3900  Ib.  ft.  The  results  given  in  the  column  h  are  obtained  by 
scaling  the  thickness  of  the  arch  ring  at  the  points  of  division. 
The  amount  of  steel  reinforcement  is  assumed  to  be  constant 
and  so  the  percentage  or  ratio  changes  inversely  as  h.  With 
these  results  scaled  and  tabulated,  the  stresses  are  found  as 
described  under  the  subject  of  combined  Flexural  and  Axial 
Stresses  pp.  90-100.  Thus,  the  compressive  stress  at  point  1  is 
larger  at  the  intrados  than  at  the  extrados  since  the  eccen- 
tricity is  on  that  side  of  the  gravity  axis.  This  stress  is 
represented  by  C  in  (81)  while  that  at  the  extrados  is  C" 
in  (82).  Since  the  reinforcement  is  symmetrically  placed  with 
reference  to  the  gravity  axis  and  not  far  from  1/10  h  from 
the  outer  edges,  Plate  VIII  may  be  used  if  the  example  falls 
under  Case  I.  From  Fig.  45  it  is  seen  that  Case  I  applies 
whenever  e/h  is  less  than  0.178  for  p  +  p'  =  0.52  per  cent,  as  is 
the  case  here.  This  means  that  the  arch  ring  at  this  section 
is  in  compression  from  intrados  to  extrados  and,  in  fact,  the 
same  is  true  of  all  parts  of  this  arch.  In  Plate  VIII,  for 
p  =  p'  =  0.0026  and  e/h  =  0.03,  N&  is  1.09  and  C  =  Ne  H/bh 
=  1.09  X  29400  -r-  (144  X  2.08)  =  107  Ib.  sq.  in.  at  the  in- 
trados. At  the  extrados  Cf  is  found  from  the  same  plate  where 


ELEMENTARY  DESIGN 


205 


Eccentricity 

Thrust  -  Pounds 

e  +  h 

Point 

Left 

Right 

Left 

Right 

Feet 

Left 

Right 

A?M 

1-16 

-0.06 

-0.32 

29400 

31280 

2.08 

0.030 

0.154 

0.52 

2-15 

-0.06 

-  0.10 

23500 

24280 

.84 

0.033 

0.055 

0.58 

3-14 

-  0.04 

+  0.10 

23500 

24280 

.60 

0.025 

0.063 

0.67 

4-13 

-0.09 

+  0.15 

21100 

21360 

.46 

0.062 

0.103 

0.74 

5-12 

-0.09 

+  0.24 

20169 

20160 

.34 

0.067 

0.179 

0.81 

6-11 

-  0.10 

+  0.20 

19600 

19500 

.20 

0.083 

0.167 

0.90 

7-10 

-0.08 

+  0.13 

19300 

19200 

1.16 

0.069 

0.102 

0.95 

8-9    • 

-  0.00 

+  0.08 

19120 

19050 

1.14 

0 

0.104 

0.98 

Spring 

-0.18 

-0.50 

29400 

31280 

2.17 

0.830 

0.230 

0.50 

AV  is  0.77  and  C"  =  0.77  X  29400  -f  (144  X  2.08)  =  76  pounds 
per  square  inch.  In  a  similar  manner  the  stresses  at  the  intrados 
and  extrados  are  computed  for  other  points  on  the  arch  and 
inserted  in  the  table.  The  positive  sign  indicates  tension  and 
the  negative  sign  compression.  The  numbers  in  the  column 
headed,  for  instance,  2—15  apply  to  either  point  2  or  15 
according  to  the  position  of  the  live  load.  The  position  of  the 
line  of  thrust  with  reference  to  the  gravity  axis  determines 
whether  the  maximum  stresses  are  on  the  intrados  or  the 
extrados. 

Temperature  Stresses.  The  range  of  temperature  in  the 
interior  of  the  arch  ring  is  somewhat  indefinite.  If  the  roadway 
be  supported  on  columns  which,  in  turn,  rest  upon  the  arch, 
leaving  the  same  exposed  on  all  sides,  the  range  is  evidently 
greater  than  when  the  spandrel  filling  is  solid.  For  the 
problem  now  being  discussed  it  will  be  assumed  that  the 
temperature  varies  between  25°  Fahr.  above  and  the  same 
amount  below  the  normal.  As  stated  in  a  previous  chapter 
the  coefficient  of  expansion  may  be  taken  as  0.0000055  for 
concrete. 

The  thrust  due  to  temperature  changes  is  given  by  formula 
(144)  and,  making  the  proper  substitutions,  it  is  found  numer- 
ically. From  (6)  Fig.  79  the  values  of  y  are  found  to  be  10.12, 
7.22,  5.00,  3.24,  1.92,  0.96,  0.32,  and  0.02  feet  for  subscripts  1  to  8. 
It  will  be  noted  that  these  values  of  y  are  measured  downward 
from  the  horizontal  line  through  the  highest  load  points  on  the 
arch  rather  than  upward  from  the  chord.  The  mean  value  of 


206 


REINFORCED  CONCRETE 


. 

CO 

o 

o 

o 

T* 

•* 

X 

rH 

^ 

vP 

co 

| 

H 

1 

I 

1 

1 

00 

- 

§ 

2 

§ 

o 

o 

S 

d 

i—  t 

1 

I 

1 

aj 

CO 

§ 

o 

§ 

S 

§ 

o 

N 

1 

l 

1 

1 

CO 

I 

. 

5! 

S 

S 

o 

o 

S 

*^* 

S 

i-H 

d 

1 

i 

1 

*o 

n< 

o 

CO 

IQ 

C*4 

-*^ 

C5 

^ 

•<11 

iO 

3 

w 

I 

1 

1 

1 

1  ' 

. 

00 

•^1 

CO 

CO 

0 

5 

S 

>2 

1 

1 

1 

j 

3 

05 

0 

S 

S 

CD 

£j 

w 

1 

1 

1 

1 

1 

^ 

% 

O 

S 

o 

o 

g 

» 

1 

1 

1 

. 

CD 

>- 

c: 

-r 

i? 

S 

CO. 

X 

W 

I 

1 

1 

1 

r"H 

1 

^ 

. 

" 

ct 

Jj| 

o 

o 

g 

a 

^H 

1 

1 

1 

S 

1 

1 

co 

7 

CO 

1 

S 

o 

1 

1 

CO 

^ 

§ 

00 

O 

CO 

z 

g 

"-1 

1 

1 

1 

\ 

. 

CO 

•* 

S 

0 

0 

e 

X 

12 

W 

1 

1 

1 

1 

(N 

«i 

S 

:~; 

"o 

§ 

S 

S 

*S 

1 

i 

1 

1 

^ 

-r 

7. 

5 

S 

2 

S 

1 

1 

I 

1 

. 

g 

00 

O 

~ 

CO 

g 

a 

1 

1 

\ 

1 

•si 

i 

3 

.i 

~ 

a 

TJ 

u< 

fa 

J| 

2 

p 

e 

1 

1 

Live  and  D 

Temperatu 

Temperatu 

Shortening 

Maximum 

I 

|.2 

8* 

«& 

P 

^  .a 

fs 


ELEMENTARY  DESIGN 


207 


ds/I  is  found  from  the  previous  work  to  be  9.34.     Then,  n  being 
the  number  of  divisions  in  half  the  arch, 

8  X  25  X  51.74  X  0.0000055     288000000 


2  (8  X  194.75  -  829.44) 


9.34 


1210  pounds. 


Also  Mc  =  -  1210  X  28.8  -f-  8  =  4350  Ib.  ft.  The  positive  sign  is 
used  when  the  temperature  is  above  and  the  negative  sign  when 
it  is  below  the  normal.  The  thrust  is  applied  at  the  mean  of  y,  = 
3.60,  below  the  crown,  and  the  eccentricity  of  application  is  the 
difference  between  this  value  and  y.  These  distances  are  given  in 
the  second  column  of  the  following  table.  The  thickness  of  the 
arch  ring  is  h  so  e/h  is  known  and  k/h  is  taken  from  Plate  X.  The 
value  of  k  is  then  found  and  C,  the  stress  on  the  edge  of  the  arch 
ring  which  is  in  the  greater  compression  is  determined  from  (92). 
If  the  thickness  of  the  arch  be  measured  radially,  the  thrust  is 
the  component  of  Hc  normal  to  the  section  considered.  Thus 
at  points  1  and  16  the  normal  component  of  Hc  is  850  pounds. 


Point 

Eccentricity 
Feet 

h  ft, 

•e,'h 

k/h 

fcin. 

*» 

1 

-6.52 

2.08 

3.13 

0.240 

6.00 

36.0 

2 

-3.62 

1.84 

1.97 

0.270 

6.20 

38.4 

3 

-  1.40 

1.60 

0.87 

0.340 

6.54 

42.5 

4 

0.36 

1.46 

0.25 

0.830 

14.50 

210.3 

5 

1.68 

1.34 

1.25 

0.330 

5.32 

28.30 

6 

2.64 

1.20 

2.20 

0.305 

4.32 

18.66 

7 

3.28 

1.16 

2.73 

0.295 

4.10 

16.81 

8 

3.58 

1.14 

3.14 

0.295      • 

4.05 

16.40 

For  these  values  of  p  and  e/h  it  will  be  found  from  Fig.  45  that 
one  edge  of  the  arch  ring  is  in  tension  at  every  point.  At  point 
1,  C  =  (2  X  850  X  6)  +  [12  X  36  +  30  X  0.0026  X  25  (12  -  25) 
12]  =  80  Ib.  sq.  in.  The  tension  in  the  steel  at  the  other  edge 
will  be,  from  (96),  3300  Ib.  sq.  in.  The  stresses  at  other  points 
are  found  in  similar  manner  and  inserted  in  the  table  on 
page  206. 

In  Fig.  84  the  half  arch  ring  is  represented  as  being  fixed  at 
the  springing  line  and,  at  the  crown,  by  the  forces  exerted  by 
the  other  half  of  the  arch.  The  thrust,  H,  due  to  change  of 
temperature,  acts  outward  in  case  of  a  rise  and  inward  in  case 
of  a  fall,  on  the  two  cantilevers  above  and  below  the  line  kk. 


208  REINFORCED  CONCRETE 

With  this  consideration  in  mind  there  can  be  no  ambiguity  as 
to  the  nature  of  the  stresses  resulting  from  temperature  changes 
alone.  Accordingly  the  temperature  stresses  at  all  points  are 

computed  for  both  intrados  and 
extrados    and    the    results    are 
^    placed    in   the   proper   columns 
~~  of  the  table,  page  206. 

Stresses  Due  to  Shortening. 
The     average    compression     in 
the  arch  ring  is   somewhat   in- 
84-  definite  owing  to  several  causes, 

notably  the  eccentricity  of  the  points  of  application  of  the  thrust. 
In  this  case  the  thrust  at  each  point  is  found  from  the  force 
polygon  and  divided  by  the  areas  of  the  sections  of  the  arch  ring 
at  the  corresponding  points.  An  average  of  110  Ib.  per.  sq.  in. 
seems  to  be  a  reasonable  value  of  the  compression  which  produces 
a  shortening  equal  to  that  produced  by  a  fall  of  t'  degrees.  Then 
as  I'd  =  PI  +  AE,  t'  =  110  +  (2,000,000  X  0.0000055)  =  10 
degrees  or  40  per  cent,  of  25  degrees.  Hence  the  stresses  due 
to  arch  shortening  are  40  per  cent,  of  those  due  to  fall  of  tempera- 
ture and  may  thus  be  computed  from  values  of  the  latter  in  the 
table  of  final  stresses. 

Final  Tension  and  Compression  Stresses.  The  dead  load 
stresses  and  those  due  to  arch  ring  shortening  are  always  active, 
and  stresses  due  to  rise  and  fall  of  temperature  are  to  be  chosen 
to  give  maximum  tension  and  compression  at  intrados  and 
extrados.  Thus  at  1  or  16  the  maximum  compression  at  the 
intrados  is  —  183  —  37  —  80  =  300  Ib.  sq.  in.  The  maximum 
compression  at  the  extrados  is  —  76  —  80  =  —  156  Ib.  sq.  in.  The 
stresses  at  other  points  are  found  in  a  similar  manner.  The 
stresses  in  the  steel  are  low  and  may  be  found  by  formula  (96) . 
It  will  be  noted  that,  in  an  arch  of  this  variety,  the  temperature 
stresses  are  very  important  and  that  in  sections  near  the  crown 
they  are  greater  than  those  due  to  the  live  and  dead  loads. 
If  the  arch  be  of  large  span,  as  100  feet,  the  live  load  should 
be  placed  in  several  positions  and  the  stresses  found.  Usually 
four  positions  are  sufficient  to  give  maximum  stresses.  Profes- 
sor Wm.  Cain  suggests  1  a  trial  of  live  loads  over  three,  five, 
six,  and  ten  tenths  of  the  span.  A  more  exact  method  is 
1  Trans.  Amer.  Soc.  C.  E.,  Vol.  IV,  pp.  191-193. 


ELEMENTARY  DESIGN  209 

described  by  the  same  author,  and  others,  by1  which  a  single 
load  is  placed  successively  at  all  division  points,  the  effect 
noted,  and  all  possible  combinations  used  to  find  the  maximum 
stresses. 

Another  valuable  discussion  of  possible  combinations  of 
stresses  in  arches  is  given  by  Mr.  A.  H.  Fuller  in  Trans.  Amer. 
Soc.  C.E.  in  connection  with  the  paper  mentioned  above. 

Shearing  Stresses.  If  the  direction  of  the  line  of  pressure 
be  different  from  that  of  the  gravity  axis  the  radial  component 
of  the  resultant  pressure  is  the  shear  in  the  section.  At  point 
16  the  thrust  is  31,280  pounds  and  by  scale  the  radial  component 
is  found  to  be  f  of  this,  or  3910  pounds.  The  area  at  this  point 
is  25  X  12  =  300  sq.  in.,  and  the  shear  is  13  pounds  per  square 
inch.  Stirrups  or  diagonal  tension  rods  are  sometimes  inserted 
for  the  double  purpose  of  holding  the  main  reinforcement  in 
place  and  also  to  prevent  distortion  due  to  excessive  concentrated 
loads. 

Final  Dimensions.  An  inspection  of  the  table  of  stresses 
shows  that  the  preliminary  assumptions  as  to  thickness  of  the 
arch  at  the  crown  and  at  the  springing  were  about  right.  The 
appearance  is  somewhat  improved  by  a  curve  of  smaller  radius 
near  the  springing  line.  The  reinforcement  will  be  furnished 
by  one  inch  rods  spaced  six  inches  center  to  center. 

Abutments.  In  arches,  more  than  with  any  other  class  of 
structures,  it  is  essential  that  the  foundations  be  unyielding. 
The  elastic  theory  is  applied 
upon  this  assumption  and  every 
effort  is  made  to  render  the  hy- 
pothesis justifiable.  Abutments 
may  fail  (a)  on  account  of  too 
great  uniform  pressure,  (6)  on 
account  of  improper  distribution 
of  the  pressure  through  crushing 
or  overturning  and  (c)  by  sliding. 

In  Fig.  85  the  thrust,  16,  is 
transmitted  by  the  arch  to  the 
amount  of  31,280  pounds.  This  is  combined  with  the  weight,  v, 
of  the  abutment  with  the  earth  filling  over  it  and  the  thrust,  h, 
of  the  earth  below  the  springing  lines,  if  the  ground  be  soft,  to 

1  "Theory  of  Solid  and  Braced  Elastic  Arches."      D.  Van  Nostrand  Co. 


210  REINFORCED  CONCRETE 

form  the  resultant  R.  In  this  case  it  is  assumed  that  the  foun- 
dation is  on  hard  pan  capable  of  sustaining  a  load  of  5  tons  per 
square  foot.  The  distance  from  the  roadway  to  c  is  scaled, 
the  weight  of  earth  filling  is  computed,  the  live  load  for  length 
ca  is  found  and  both  are  added  to  the  weight  of  an  assumed 
area  of  concrete  abc  to  produce  v.  The  weight  so  found  is  com- 
bined with  the  maximum  thrust  of  the  arch  ring,  as  at  load  16 
of  the  force  polygon,  Fig.  79.  The  maximum  unit  pressure 
due  to  R  is  found  as  in  the  case  of  retaining  walls,  and  must  not 
exceed  5  tons  per  square  foot.  The  details  of  this  abutment 
design  are  proposed  as  a  separate  problem  for  the  student. 

Symmetrical  Proportions.  Attention  is  called  to  Fig.  81 ,  which 
shows  a  reinforced  concrete  bridge  at  Austin,  Texas.  This 
structure  was  designed  by  Waddell  and  Harrington,  and  illus- 
trates the  possibility  of  combining  pleasing  details  with  grand 
proportions. 

Sequence  of  Work  in  Arch  Design 

General  Instructions.  Size  of  drawing  sheet  is  22"  X  30"; 
border,  20"  X  28".  The  scale  of  the  arch  ring  is  1  inch  =  2  feet, 
that  of  the  force  polygon  is  1  inch  =  2000  pounds.  The  arch 
ring  is  at  the  top  of  the  sheet,  the  force  polygon  is  at  the  right, 
and  the  trial  equilibrium  polygon  is  below  the  arch  ring. 

Data.     Clear  span, ;  rise, ;  C  = ;  St  =  -  -  Ib.  sq. 

in.;  shear  in  concrete  alone,  -  -  Ib.  sq.  in.;  with  diagonal  rein- 
forcement, -  -  Ib.  sq.  in.;  bond,  -  -  Ib.  sq.  in.;  n  =  15  Ec  = 

2,000,000  inches4;   per  cent,   of   steel   at   crown  = .     Dead 

load:  concrete, Ib.  cu.  ft.;  earth  filling, Ib.  cu.  ft.;  ballast, 

paving,  etc., Ib.  cu.  ft. 

Live  load: Ib.  per  sq.  ft.  Assume  the  live  load  as  extend- 
ing over  one-half  of  the  arch  ring  for  spans  up  to  50  feet.  For 
long  spans  the  live  load  should  also  be  tried  over  T3o  and  T°cr  of 
the  span. 

The  loads  are  considered  to  act  vertically  unless  the  rise  of 
the  arch  is  about  J  of  the  span. 

Design.  (1)  Determine  the  thickness  of  the  arch  ring  at  the 
crown  by  Weld's  formula:  t  =  Vs  +  O.ls  +  0.005  wt  +  0.0025 
wd  in  inches. 

(2)  Lay  out  the  arch  ring  of  the  clear  span  and  rise,  using  either 
segmental  circular,  a  three  or  a  five-centered  curve  for  the 


ELEMENTARY  DESIGN  211 

intrados.  Use  a  segment  of  a  circle  for  the  extrados.  The 
thickness  at  the  springing  line  is  from  2  to  2.5  that  at  the  crown, 
at  the  quarter  points  it  is  about  1.3  that  at  the  crown. 

(3)  Draw  the  gravity  axis  midway  between  the  extrados  and 
intrados.     This  curve  may  often  be  the  segment  of  a  circle. 

(4)  Divide  the  arch  ring  by  vertical  lines  into  parts  such 
that  ds/I  is  constant.     From  8  to  20  divisions  are  usually  suffi- 
cient.    For  method  see  page  198.     The  reinforcement  is  usually 
taken,  for  trial,  as  about  one  per  cent,  of  the  area  of  a  section 
of  the  arch  ring  at  the  crown.     This  is  placed  from  li  to  2| 
inches  from  the  edges. 

(5)  Construct  a  reduced  contour  load  line  above  the  arch. 
Divide  the  area  between  the  intrados  and  the  reduced  contour 
by  vertical  lines  through  the  points  established  in  (4).     Find 
the  area  and  center  of  gravity  of  each  trapezoid  to  determine 
the  magnitude  and  point  of  application  of  the  loads.     Number 
these  loads  from  left  to  right. 

(6)  Draw  the  load  line,  assume  a  pole    opposite    the  point 
between  the  two  loads  on  either  side  of  the  crown,  and  draw 
the  rays.     (The  rays  may  be  omitted.) 

(7)  Draw  the  trial  equilibrium  polygon  and  the  closing  chord. 
Scale  the  ordinates  between  the  chord  and  the  polygon,  find 
the  mean  ordinate  and  the  eccentricity,  and  locate  k'k'  as  de- 
scribed on  page  196. 

(8)  Scale  the  ordinates  between  the  chord  and  the  gravity 
axis  in  (a),  Fig.  79,  and  locate  kk.     On  kk  plot  the  points  p  from 
the  intersection  of  the  trial  equilibrium  polygon  and  the  line 
k'k'.     Tabulate  the  ordinates  to  the  gravity  axis  and  the  trial 
equilibrium  polygon,   and  compute  the  true  pole  distance  as 
indicated  on  page  196. 

,  (9)  Draw  the  true  equilibrium  polygon  through  the  points 
p.  Scale  the  eccentricities  and  compute  the  stresses  due  to 
loads  at  the  intrados  and  extrados  in  the  manner  described  in 
Chapter  IV.  Find  the  shear  at  all  points. 

(10)  Compute  the  thrust  due  to  temperature  changes  accord- 
ing to  formula  (144). 

(11)  Compute  the  stresses  at  intrados  and  extrados  due  to 
temperature    changes.     Refer   to    Fig.    82   in   determining   the 
nature  of  these  stresses. 

(12)  Find  tf,  the  change  in  temperature  that  would  produce 


212 


REINFORCED  CONCRETE 


stresses  equal  to  those  due  to  average  compression  in  the  arch 
ring.  Find  the  stresses  due  to  shortening  of  the  arch  ring  by 
the  ratio  t  +  t'. 

(13)  Make  a  table  of  stresses  similar  to  that  on  page  206.  and  find 
the  final  maximum  and  mini  mum  stresses  at  intrados  and  extrados. 

(14)  Compare   the   final   stresses   with   those   allowable   and 
make  such  changes  in  the  thickness  of  the  arch  ring  or  in  the 
amount  of  reinforcement  as  are  necessary  to  make  the  assumed 
and  computed  stresses  agree. 

(15)  Design  an  abutment,  estimate  the  pressures  on  it,  com- 
bine its  weight  with  these  pressures,  and  find  the  maximum 
resulting  pressure  on  the  ground.     If  the  allowable  earth  resist- 
ance be  exceeded,  increase  the  width  of  the  footing  accordingly. 

Design  of  a  Beam  and  Girder  Floor 

This  design  will   not  be  worked  out  in  numerical  detail,  but 
the  sequence  of  computation  with  proper  references  to  foregoing 

pages  will  be  given.  The 
panels  will  be  assumed 
of  the  same  dimensions 
and  carrying  the  same 
load  as  the  girderless 
floor  on  page  129,  18  ft. 
8  in.  by  19  ft.  1  in., 
and  divided  as  shown 
in  Fig.  86. 

The  slabs  are  6  feet 
2f  inches  in  span,  par- 
tially fired  and  contin- 
uous, and  the  bending 
moment  may  be  taken 
as  iV  wl2.  The  thick- 
ness may  be  assumed,  to 
determine  the  load,  as  4 
inches,  so  the  live  and 
dead  load  is  225  +  J 
X  150  =  275  Ib.  sq.  ft. 
Fl«-  86.  Assume  the  allowable 

stresses  in  steel  and  concrete  to  be  16,000  and  600  Ib.  sq.  in. 
The  solution  is  made  by  means  of  Plate  I. 


Section 


Girders 


I 

i 

,    . 

h-~ 82* i  *  - 


u= 


, 


U 


Plan 


ELEMENTARY  DESIGN  213 

The  Beams.  These  are  19  ft.  1  in.  in  length,  and  carry, 
beside  their  own  load,  225  X  6  X  19.08  pounds.  The  dead  load 
may  be  assumed  at  half  the  added  load.  The  slab  forms  the 
flange  of  a  T-beam,  of  which  the  web  is  to  be  designed.  The 
section  area  is  first  determined  to  resist  the  shear,  as  explained 
in  the  design  of  the  T-beam,  page  157.  The  width  of  flange  is 
to  be  taken  in  accordance  with  the  recommendation  of  the 
Joint  Committee,  page  150.  The  amount  or  percentage  of  steel 
may  be  assumed,  and  the  width  of  the  web  made  to  conform 
with  this  assumption,  page  82.  The  negative  bending  moment 
may  be  taken  as  TV  wl2,  and  the  beam  is  rectangular  with  double 
reinforcement  rather  than  of  T-section  at  the  supports.  In  the 
computations  Plate  II  may  be  used.  It  is  possible  that 
the  section  may  have  to  be  deepened  at  the  supports  to  keep  the 
area  of  steel  reasonable.  A  new  depth  is  assumed  and  the  new 
percentage  of  steel  is  thus  fixed,  the  stresses  are  computed  and 
a  new  assumption  as  to  depth  is  made  if  necessary.  The  design 
of  diagonal  tension  members,  stirrups,  and  bent-up  rods  is  made 
as  explained  before.  The  section  at  the  support  is  investigated 
for  bond  stresses. 

The  Girders.  The  girders  have  a  span  of  18  feet  8  inches, 
and  the  loads  may  be  assumed  to  be  applied  at  the  third  points. 
The  whole  load  is  taken  for  trial,  as  150  per  cent,  of  the  applied 
load.  The  width  of  the  web  is  determined  by  the  shear,  and 
the  breadth  or  depth  may  be  assumed.  The  amount  of  steel 
is  computed  and  the  breadth  is  changed  if  necessary.  For 
economy,  the  depth  is  made  as  great  as  circumstances  will 
allow.  Negative  bending  occurs  over  the  supports,  the  T- 
section  there  becomes  rectangular,  and  -the  depth  may  be  in- 
creased for  a  short  distance.  The  computations  are  made  by 
means  of  the  formulas  for  double  reinforced  rectangular  beams 
and  Plate  IT.  Diagonal  tension  reinforcement  is  provided  for 
by  bent-up  rods  and  stirrups  as  described  before.  At  the  sup- 
ports the  sections  are  investigated  for  bond  stresses  and  changes 
are  made  in  dimensions  if  necessary.  The  weights  are  recom- 
puted, and  sections  may  be  verified  or  altered  accordingly.  In 
choosing  the  steel  the  difficulties  in  handling  large  rods  should  be 
considered,  and  on  the  other  hand  a  complexity  of  rods  interferes 
with  the  best  results  in  pouring  the  concrete.  Rods  f  inch  in  diam- 
eter are  much  used  in  tension,  and  f  and  |-inch  rods  for  stirrups. 


214 


REINFORCED  CONCRETE 


Sequence  of  Work.     Design  of  Reinforced  Concrete  Hollow  Dam 

General  Instructions.  Size  of  drawing  sheet  22"  X  30"; 
border,  20"  X  28". 

Scale  of  drawing:  1  inch  =  4  feet. 

Show  a  vertical  section  through  a  buttress  at  right  angles  to 
the  axis  of  the  dam.  Make  this  drawing  at  left  end  of  the 
sheet. 

Show  an  elevation  including  a  panel  and  two  buttresses,  look- 
ing upstream.  Draw  this  elevation  to  the  right  of  the  vertical 
section. 

Show  all  reinforcing  bars  in  both  section  and  elevation;  indi- 
cate on  the  vertical  section  the  forces  acting,  the  centers  of 

gravity  as  used,  and  diagrams  for 
pressure  upon  the  foundation. 

Data.  Elevation  of  river  bed  = 
120;  crest  of  dam  =  160;  high 
water  =  158. 

The  slope  of  the  upstream  face 
=  45°;  down-stream  face  vertical 
with  no  wall  between  buttresses. 

1  Try  a  crest  2  feet  wide;  buttresses 

2  feet  thick  at  bottom  and  1  foot 
thick  at  top;  spaced  10  feet  C  to 

k^rr  *"  "  ~  ~  —  ^-"-^*    C;   struts  between   buttresses    12 

|J 


Section  x 


FIG.  87. 


zontally  and  20  feet  apart  verti- 
cally. 

The  foundation  at  the  river-bed  elevation  is  solid  rock,  so 
no  concrete  floor  will  be  used. 

Allowable  pressure  on  foundation  15  tons  per  sq.  ft.;  weight 
of  concrete,  150  pounds  per  sq.  ft.;  C  =  500  Ib.  per  sq.  in.; 
St  =  15,000  Ib.  per  sq.  in.;  vf  =  30  Ib.  per  sq.  in.;  for  plain  con- 
crete, and  100  Ib.  for  reinforced  concrete;  bond  stress  =  80  Ib. 
per  sq.  in;  n  =  15. 

Design.  1.  Calculate  thickness  of  the  face  slab  at  the  bot- 
tom of  the  dam,  at  ele.  135,  and  at  ele.  150,  using  M  =  iV  wl2 
for  both  positive  and  negative  moments.  Consider  the  weight 
of  the  slab  in  calculating  the  maximum  bending  moment. 

2.  Test  the  thickness  of  the  slab  at  the  buttresses  for  shear. 


ELEMENTARY  DESIGN  215 

3.  Find  the  weight  of  the  concrete  in  one  panel,  including 
slab,  struts,  and  one  buttress.     Call  this  W. 

4.  Calculate  the  total  water  pressure,  both  vertical  and  hori- 
zontal, on  one  panel.     Call  these  V  and  N  respectively. 

5.  Find  the  weight  due  to  concrete  and  water  at  bottom  of 

W  -f  V 
a  buttress  on  a  strip  one  foot  wide.     This  equals ~ —  since 

buttress  is  2  feet  wid  at  bottom  (neglecting  the  component 
of  the  weight  of  the  slab  parallel  to  the  face  which  is  carried 
at  the  base  of  the  slab  between  buttresses). 

6.  Find  the  center  of  gravity  of  a  panel  of  the  concrete  referred 
to  downstream  edge.     Find  e.g.  of  slab  buttress  and  struts  sep- 
arately, then  take  moments  to  find  e.g.  of  the  panel,  that  is,  the 
point  of  application  of  W. 

7.  Find  point  of  application  of  H  and  V. 

8.  Find  center  of  gravity  of  W  and  V  combined. 

9.  Find  point  where  the  resultant  of  W  +  V  and  H  cuts  the 
base  of  the  dam  referred  to  down-stream  edge. 

10.  Find  maximum  and  minimum  pressures  per  sq.   ft.  on 

*u    f  W      ®dw 

the  foundation,  using  general  formula:  r  =  -j-  =•=  — « — . 

11.  Draw  diagram  showing  distribution  of  pressure  along  the 
base  due  to  the  combination  of  water  and  concrete. 

12.  Determine    the    stability    against    overturning,     taking 
moments  about  the  downstream  edge.     Factor  of  safety  of  3. 

13.  Determine    the    stability    of    the    dam    against    sliding 
using   0.70   as   coefficient    of  friction.     Factor  of  safety  of   1. 
(Adhesion  of  concrete,  and  toe  and  heel  walls  offer  additional 
resistance.) 

14.  Determine  the  effect  of  shearing  resistance  upon  sliding. 
(Make  heel  wall  4  feet  deep  and  wide  as  base  of  apron  slab. 
If  the  required  factors  of  safety  are  not  reached  a  new  design 
should  be  made  by  widening  the  cut  or  making  a  small  slope 
angle.) 

15.  Determine  the  actor  of  safety  against  crushing. 

16.  Choose   and   arrange   reinforcing   rods   in   the   slab,   con- 
sidering normal  stress,  shear,  bond,  and  diagonal  tension. 

17.  Put  in  horizontal  rods  to  bind  the  slab  to  the  buttresses. 
No  calculations  are  necessary  since  the  slab  is  in  compression 
against  the  buttresses. 


216  REINFORCED  CONCRETE 

18.  Reinforce  the  struts  as  columns. 

19.  Determine  how  such  reinforcement   for  the  bending  mo- 
ment due  to  the  weight  of  the  strut  is  necessary. 

20.  Compute  the  cost  per  linear  foot  of  dam.     Concrete  $£ 
per  cubic  yard  and  steel  3  cents  per  pound. 


INDEX 


Abutments  of  an  Arch,  209 
Adhesion  of  Concrete   to   Steel   (see 

Bond).  23,  31 
Aggregate,  12,  140 
Analyses 

arch,  algebraic,  183 

arch,  graphic,  191 

parabolic     assumption     55-68 

straight  line  assumption,  38 

see  chap.  IV,  37 
Anchoring  Bars,  24,  150,  167 
Ancient   Reinforced   Concrete   struc- 
tures, 1 

Angle  of  Bent-up  Bars,  83 
Areas  of  Rods  and  Bars,  table  156 
Arches 

abutments,  209 

analysis,   algebraic   183,   graphic 
191 

dead  loads,  197 

depth  of  keystone,  197 

earth  pressure,  197 

example  of  design,  197 

extrados,  198 

formulas,  general,  184,  185 

formulas  for  thrust,  moment  and 
shear  at  crown,  189 

hingeless,   187 

intrados,  198 

live  loads,  197 

moment  at  crown,  189 

ring  shortening,  191 

steel  reinforcement,  198 

stresses  in  ring,  189 

temperature  stresses,  189 

three-hinge  arch,  186, 

thrust  at  crown,  189 

two-hinge  arch,  186 
Axial  loads  90-101 


Bag  of  portland  cement,  weight  of,  140 
Bands  in  columns,  111,  116 
Barrel  of  portland  cement,  140 
Bars,  deformed,  24 

depth    of    concrete    below,    144 
length  to  develop  tensile  strength, 

80 

points  of  bending  up,  77 
spacing  of,  77,  82 
Table  of  weights,  areas  and  peri- 
meters, 156 

217 


Beams 

analyses,  39,  146 

axial  loads,  90 

continuous,  122,  123,  124,  149 

diagonal  tension,  71 

diagrams   for    (see    Diagrams) 

double  reinforcement,  44 

economic  proportions,  132 

formulas  (see  Formulas) 

homogeneous,  69 

horizontal  shear,  69 

manner  of  failures,  29,  74 

neutral  axis,  26 

parabolic  analysis,  55-59 

repetition  of  loads,  32 

shear,  17 

slab  loads  on,  126 

spacing  of  tension  bars,  82 

straight  line  analysis,  39, 

T-beam  (see  T-beams) 

tension  failures,  29 

vertical  shear,  68 

web  stresses,  76 
Bond  of  concrete  to  steel,  22 

formula  for,  25 

mechanical,  24 

stresses  in  retaining  walls,  176 

tests  for,  31 

working  stress,  24,  154 
Broken  stone,  specifications,  12 

amount  in  cu.  yd.  concrete,  35 

Cement,    approximate    quantities    in 

cu.  yd.  concrete,  35 
barrel,  volume,  140 
barrel,  weight,  140 
cost,  ratio  diagram,  137 
in  cu.  yd.  concrete,  35 
production  in  U.  S.,  3 
specifications,  6 
tests,  6 

Cinder  concrete,  85 
Columns, 

effective  area  of,  151 

failures,  diagrams  of,  108-109, 110 

formulas  for  103,   111,  117,  120, 

121 
hooped  or  spiral    reinforcement, 

112,  113 

increase  in  strength   due   to   re- 
inforcement, 104 


218 


INDEX 


longitudinal    reinforcement,    107 

strength  of  plain  columns  105,106 

strength   of  reinforced    columns, 
112,  113 

Tests  of  plain,  105,  106 

Tests,  Considered,  118 

Tests,  Univ.  of  Illinois,  105 

Tests,  Watertown,  119 

Use  of,  102 

working  rules  for,  152 
Compression  strength  of  concrete,  16 

working  strength  of  concrete,  17 
Concrete,  bond  with  steel  23,  31 

cinder,  85 

consistency,  141 

crushing  strength,  16 

elasticity,  19,  20 

expansion,  86 

fatigue  of,  31,  32 

fireproofing,  84,  144 

heat  conductivity,  86 

ingredients  in  a  cubic  yard,  35 

mixing,  140 

placing,  141 

relation  to  aggregate,  140 

shrinkage,  143 

strength  of,  14,  15,  16,  17,  18,  19, 
23,  31 

uses,  of  4 

waterproofing,  145 

working  strength,  16,  17,  19,  24, 

153,  154,  155 

Consistency  of  concrete,  7,  141 
Continuous  beams,  122,  123,  124,  149 

spans,  147 

Cracks  in  walls  due  to  temperature,  87 
Crathorne,  A.  R.,  122 

Dam,  design  of  hollow,  214 
Diagonal  tension,  68,  154 
Diagram  of  sieve  analysis,  9 

stress-strain,  21,  32,  38,  58,  114, 

115 

stresses  in  beams,  37 
for  rectangular  beams,  41 
double  reinforced  beams,  47 
for  T-beams,  51 
full  parabolic  relation,  59 
general  parabolic  relation,  63,  65 
k-j-d  in  terms  of  q,  67 
stresses  in  a  homogeneous  beam, 

72 

of  points  for  bending  up  bars,  78 
of  distances  between   cracks,   88 
for    stresses    under    flexure    and 
axial  loads,   93,  94,  95,  97,  99 
showing  ratio  of  strength  of  re- 
inforced to  plain  columns,  104 
for    determining    economic    pro- 


portions in  beams,  135 
showing  ratio  of  cost  of  steel  to 

that  of  concrete,  137 
Double  reinforcement  of  beams,  44 

Earth,  bearing  power  of,  169 
horizontal  pressure  of,  165 
weight  of,  166 

Economic  proportions  of  beams,   132 
Elastic  deformation  of  steel,  35 

limit  of  steel,  35 
Equilibrium  polygon,  trial,  202 
Equilibrium  polygon  true,  203 
Expansion   and    contraction   of   con- 
crete, 32 

Fatigue  of  concrete,  31,  32 
Fineness  of  concrete,  6 
Fireproofing  of  concrete,  84,  85,  144 
Fires,  effect  of,  on  concrete,  85 
Flexural  tests  of  beams,  25 
Flexural  and  axial  loads  on  beams, 
90-101 

axial  compression,  92-95 

axial  tension,  101 

diagrams  for,  93,  94,  95,  97,  99 
Floor  slabs,  124-127 
Floors,  beam  and  girder  design,  212 

girderless,  129 

loads  from  slabs,  127 
Forms,  142,  171,  173 
Formulas,  bond,  25 

diagonal  tension,  71 

double  reinforcement,  44 

flexural,  39,  40 

flexural  and  axial  loads,  93-98 

for  arch  analysis,  184-191 

for   economic  designs    of    beam, 
132,  134 

for  columns,   103,  111,  116,   121 

T-beams,  49,  53 

ultimate  loads,  56-61 

web  stresses,  76 
Freezing    weather,     laying    concrete 

in,  142 

Frictional  resistance  of  bars,  22 
Fuller's  rule  for  proportioning  con- 
crete, 35 

Girders  (see  Beams) 
Gravel,  coarseness  of,  12 


Heat  conductivity  of  concrete,  86 

Hingeless  arch,  187 

Historical  sketch,  1 

Hooke's  law,  37 

Hooks,  for  anchoring,  24,  150 


INDEX 


219 


Hooped  columns,  109-118 

Live  loads,  on  highways,  197 
assumptions  for  design,  157 

Mechanical  bond,  24 

Merriman's  formulas  for  continuous 

beams,  122 
Moments,  due  to  eccentric  loads,  91 

in  an  arch  ring,  188 
Mortar,  retempering,  141 
Mushroom  system  of  floors,  129 

Neutral  axis,  26 
Nomenclature,  general,  38 

Parabolic    theory    of   stresses,    55-67 
Perimeters  of  rods  and  bars,  156 
Pressure  of  earth,  165 
Placing  concrete,  141 

in  freezing  weather,  142 
Plates  (see  Diagrams) 
Portland    cement   specifications,    6 
Proportioning  concrete,  34,  35 

Quantities,    ingredients    in    cu.    yd. 
concrete,  35 

Reinforced    concrete,    elongation,    32 

fire-resisting    qualities,    84,    85 

history  of,  1 

repeated  loads,  31,  32 

shrinkage,  33 

temperature  stresses,  189,   205 

uses  of,  4 

working  stresses,  bond,  24,  31, 154 
compression,  17,   154 
diagonal  tension,  150 
shear,   19,  31,   150 
tension,  14 
Retaining  Walls,    design,    164,  174 

anchoring    reinforcement,    171 

bond  stresses,  170 

footing,  167,  177 

forms,  142,  171,  173 

reinforcement,    166,    174,    180 

shear  at  footing,  166 

sequence  of  computation,  171,  181 

thrust  of  earth,  165 
Rods  (see  Bars) 
Rubble  concrete,  142 

Sand,  easy  tests  of,  11 

effect  of  good  and  bad,  11 
relation  to  broken  stone,  140 
specifications,  7 

Shear  definition,  17 

strength  of  concrete  in,  18 
working  stresses,  19,  154 


Shrinkage  of  concrete  during  construc- 
tion, 143 

reinforcement  for,  152 
Sieve  analysis  of  sand,  9 
Specifications  for,  aggregates,  12 

cement,  6 

mixing,  140 

sand,  7 

steel  reinforcement,  34 
Slab,  diagram  lor  solution  of,  162 

load  distribution    to  beams,  126 

reinforcement  for,  148 

supported  on  four  sides,  125 
Spacing  of  bars,  77,  149,  156 

of  stirrups,  76,  151,  160 

rule  for,  82 
Surface  finish,  145 
Steel,  coefficient  of  elasticity,  34,  155 

coefficient    of    expansion,     86 

corrosion  of,  84 

specifications  for,  34 

strength  of,  34 

working  strength  of,  155, 166, 176 
Stirrups,  bond  stresses  in,  80 

spacing,  76,  151,  160 

stresses  in,  80 

table  for  spacing,  160 
Stress-strain  diagram,  21 
Stresses,    assumptions    for    computa- 
tions, 147 

Tables  for  analysis  of  sand,  10 
areas  of  rods,  156 
bond   strength   of   concrete    and 

steel,  23,  31 

coefficient    of   elasticity   of    con- 
crete, 20 
columns,   strength  of,    106,    107, 

112,  113,  118,.  119 
compressive  strength  of  concrete, 

16,  18,  19 

concrete,  ingredients  in  cu.  yd.,  30 
determination  of  neutral  axis,  27, 

28 
distribution  of  loads  from  slabs 

to  beams,  125 

location  of  k'  k'  in  an  arch,  194 
perimeters  of  rods,  156 
production  of  concrete  in  U.  S.,  3 
shearing  strength  of  concrete,  18, 

19,  31 

spacing  of  rods,  156 
strength  of  steel,  34 
stresses  in  girderless  floors,  130 
stirrups,  spacing  of,  160 
T-beams;  strength,  of  52 
tensile  strength  of  concrete,  14, 15 
weight  of  bars  and  rods,  156 
Talbot,  Prof.  A.  N.  tests  by,  41 


220 


INDEX 


T-beam,  analysis  of  stresses,  48 

design,  155 

relation  of  floor  slab  to  flange,  148 

strength  of,  52 

Temperature  stresses,  in  arches,  189, 
205 

reinforcement  for,  88,  89 
Tests  (see  Bond,  Compression,  Shear, 

etc.) 
Theory  of  reinforced  concrete  beams, 

assumptions,  147 
Three-hinged  arch,  186 
Thrust  in  an  arch  ring,   stresses  due 

to,  191 
Two-hinged  arch,  186 

Voids,  classification  of  sand  by,  10 


Walls  (see  Retaining  Walls) 
Waterproofing,  145 
Web  reinforcement,  24,  75-83,  150, 151 
Weight  of  bars,  156 

bag  cement,  140 

barrel  cement,  140 

cubic  foot  of  earth,  166 

reinforced  concrete  per  cu.  ft.,  33 

rods,  156 

sand  per  cu.  ft.,  10 
Working  stresses  columns,  152 

concrete,  bond,  24,  154 

compression,   17,  154 

shear,  19,  154 

tension,  16 

for  steel,  155 


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